Probability Unit 6 Problem : A spinner has 4 equal sectors colored yellow, blue, green and red. What are the chances of landing on blue after spinning the spinner? What are the chances of landing on red? Solutio n: The chances of landing on blue are 1 in 4, or one fourth. The chances of landing on red are 1 in 4, or one fourth. This problem asked us to find some probabilities involving a spinner. Let's look at some definitions and examples from the problem above. Definition Example An experiment is a situation involving chance or probability that leads to results called outcomes. In the problem above, the experiment is spinning the spinner. An outcome is the result of a single trial of an experiment. The possible outcomes are landing on yellow, blue, green or red. An event is one or more outcomes of an experiment. One event of this experiment is landing on blue. Probability is the measure of how likely an event is. The probability of landing on blue is one fourth. In order to measure probabilities, mathematicians have devised the following formula for finding the probability of an event. Probability Of An Event P(A) = The Number Of Ways Event A Can Occur The total number Of Possible Outcomes
Transcript
Probability Unit 6
Problem: A spinner has 4 equal sectors colored yellow, blue, green and red. What are the chances of landing on blue after spinning the spinner? What are the chances of landing on red?
Solution: The chances of landing on blue are 1 in 4, or one fourth. The chances of landing on red are 1 in 4, or one fourth.
This problem asked us to find some probabilities involving a spinner. Let's look at some definitions and examples from the problem above.
Definition ExampleAn experiment is a situation involving chance or probability that leads to results called outcomes.
In the problem above, the experiment is spinning the spinner.
An outcome is the result of a single trial of an experiment.
The possible outcomes are landing on yellow, blue, green or red.
An event is one or more outcomes of an experiment. One event of this experiment is landing on blue.
Probability is the measure of how likely an event is. The probability of landing on blue is one fourth.
In order to measure probabilities, mathematicians have devised the following formula for finding the probability of an event.
Probability Of An Event
P(A) = The Number Of Ways Event A Can OccurThe total number Of Possible Outcomes
The probability of event A is the number of ways event A can occur divided by the total number of possible outcomes. Let's take a look at a slight modification of the problem from the top of the page.
Experiment 1: A spinner has 4 equal sectors colored yellow, blue, green and red. After spinning the spinner, what is the probability of landing on each color?
Outcomes: The possible outcomes of this experiment are yellow, blue, green, and red.
Probabilities: P(yellow) =
# of ways to land on yellow = 1
total # of colors 4
P(blue) = # of ways to land on blue = 1total # of colors 4
P(green) = # of ways to land on green = 1total # of colors 4
P(red) = # of ways to land on red = 1total # of colors 4
Experiment 2: A single 6-sided die is rolled. What is the probability of each outcome? What is the probability of rolling an even number? of rolling an odd number?
Outcomes: The possible outcomes of this experiment are 1, 2, 3, 4, 5 and 6.
Probabilities:
P(1) = # of ways to roll a 1
= 1
total # of sides 6
P(2) = # of ways to roll a 2
= 1
total # of sides 6
P(3) = # of ways to roll a 3
= 1
total # of sides 6
P(4) = # of ways to roll a 4
= 1
total # of sides 6
P(5) = # of ways to roll a 5
= 1
total # of sides 6
P(6) = # of ways to roll a 6
= 1
total # of sides 6
P(even) =
# ways to roll an even number = 3 = 1
total # of sides 6 2
P(odd) = # ways to roll an odd number = 3 = 1total # of sides 6 2
Experiment 2 illustrates the difference between an outcome and an event. A single outcome of this experiment is rolling a 1, or rolling a 2, or rolling a 3, etc. Rolling an even number (2, 4 or 6) is an event, and rolling an odd number (1, 3 or 5) is also an event.
In Experiment 1 the probability of each outcome is always the same. The probability of landing on each color of the spinner is always one fourth. In Experiment 2, the probability of rolling each number on the die is always one sixth. In both of these experiments, the outcomes are equally likely to occur. Let's look at an experiment in which the outcomes are not equally likely.
Experiment 3: A glass jar contains 6 red, 5 green, 8 blue and 3 yellow marbles. If a single marble is chosen at random from the jar, what is the probability of choosing a red marble? a green marble? a blue marble? a yellow marble?
Outcomes: The possible outcomes of this experiment are red, green, blue and yellow.
Probabilities: P(red) =
# of ways to choose red =
6 =
3
total # of marbles 22 11
P(green) = # of ways to choose green
= 5
total # of marbles 22
P(blue) = # of ways to choose blue
= 8
= 4
total # of marbles 22 11
P(yellow) = # of ways to choose
yellow = 3
total # of marbles 2
2
The outcomes in this experiment are not equally likely to occur. You are more likely to choose a blue marble than any other color. You are least likely to choose a yellow marble.
Experiment 4: Choose a number at random from 1 to 5. What is the probability of each outcome? What is the probability that the number chosen is even? What is the probability that the number chosen is odd?
Outcomes: The possible outcomes of this experiment are 1, 2, 3, 4 and 5.
Probabilities: P(1) = # of ways to choose a 1 = 1total # of numbers 5
P(2) = # of ways to choose a 2 = 1total # of numbers 5
P(3) = # of ways to choose a 3 = 1total # of numbers 5
P(4) = # of ways to choose a 4 = 1total # of numbers 5
P(5) = # of ways to choose a 5 = 1total # of numbers 5 P(even) = # of ways to choose an even number = 2total # of numbers 5
P(odd) = # of ways to choose an odd number = 3total # of numbers 5
The outcomes 1, 2, 3, 4 and 5 are equally likely to occur as a result of this experiment. However, the events even and odd are not equally likely to occur, since there are 3 odd numbers and only 2 even numbers from 1 to 5.
Summary: The probability of an event is the measure of the chance that the event will occur as a result of an experiment. The probability of an event A is the number of ways event A can occur divided by the total number of possible outcomes. The probability of an event A, symbolized by P(A), is a number between 0 and 1, inclusive, that measures the likelihood of an event in the following way:
If P(A) > P(B) then event A is more likely to occur than event B. If P(A) = P(B) then events A and B are equally likely to occur.
Exercises
Directions: Read each question below. Select your answer by clicking on its button. Feedback to your answer is provided in the RESULTS BOX. If you make a mistake, choose a different button.
1. Which of the following is an experiment?
Tossing a coin.
Rolling a single 6-sided die.
Choosing a marble from a jar.
All of the above.
RESULTS BOX:
2. Which of the following is an outcome?
Rolling a pair of dice.
Landing on red.
Choosing 2 marbles from a jar.
None of the above.
RESULTS BOX:
3. Which of the following experiments does NOT have equally likely outcomes?
Choose a number at random from 1 to 7.
Toss a coin.
Choose a letter at random from the word SCHOOL.
None of the above.
RESULTS BOX:
4. What is the probability of choosing a vowel from the alphabet?
None of the above.
RESULTS BOX:
5. A number from 1 to 11 is chosen at random. What is the probability of choosing an odd number?
None of the above.
RESULTS BOX:
Certain and Impossible Events
Unit 6
Experiment 1: A spinner has 4 equal sectors colored yellow, blue, green, and red. What is the probability of landing on purple after spinning the spinner?
Probability: It is impossible to land on purple since the spinner does not contain this color.
P(purple) = 0 = 04
Experiment 2: A teacher chooses a student at random from a class of 30 girls. What is the probability that the student chosen is a girl?
Probability: Since all the students in the class are girls, the teacher is certain to choose a girl.
P(girl) = 30 = 130
In the first experiment, it was not possible to land on purple. This is an example of an impossible event. In the second experiment, choosing a girl was certain to occur. This is an example of a certain event.
The next experiment will involve a standard deck of 52 playing cards, which consists of 4 suits: hearts, clubs, diamonds and spades. Each suit has 13 cards as follows: ace, deuce, three, four, five, six, seven, eight, nine, ten, jack, queen, and king. Picture cards include jacks, queens and kings. There are no joker cards. There are only 4 of a kind, for example, 4 tens.
Experiment 3: A single card is chosen at random from a standard deck of 52 playing cards. What is the probability that the card chosen is a joker card?
Probability: It is impossible to choose a joker card since a standard deck of cards does not contain any jokers. This is an impossible event.
P(joker) = 0 = 052
Experiment 4: A single 6-sided die is rolled. What is the probability of rolling a number less than 7?
Probability: Rolling a number less than 7 is a
certain event since a single die has 6 sides, numbered 1 through 6.P(number < 7) = 6 = 16
Experiment 5: A total of five cards are chosen at random from a standard deck of 52 playing cards. What is the probability of choosing 5 aces?
Probability: It is impossible to choose 5 aces since a standard deck of cards has only 4 of a kind. This is an impossible event.
P(5 aces) = 0
= 052
Experiment 6: A glass jar contains 15 red marbles. If a single marble is chosen at random from the jar, what is the probability that it is red?
Probability: Choosing a red marble is certain to occur since all 15 marbles in the jar are red. This is a certain event.P(red) = 15 = 115
Summary: The probability of an event is the measure of the chance that the event will occur as a result of the experiment. The probability of an event A, symbolized by P(A), is a number between 0 and 1, inclusive, that measures the likelihood of an event in the following way:
If P(A) > P(B) then event A is more likely to occur than event B. If P(A) = P(B) then events A and B are equally likely to occur. If event A is impossible, then P(A) = 0. If event A is certain, then P(A) = 1.
Exercises
Directions: Read each question below. Select your answer by clicking on its button. Feedback to your answer is provided in the RESULTS BOX. If you make a mistake, choose a different button.
1. A glass jar contains 5 red, 3 blue and 2 green jelly beans. If a jelly bean is chosen at random from the jar, then which of the following is an impossible event?
Choosing a red jelly bean.
Choosing a blue jelly bean.
Choosing a yellow jelly bean.
None of the above.
RESULTS BOX:
2. A spinner has 7 equal sectors numbered 1 to 7. If you spin the spinner, then which of the following is a certain event?
Landing on a number less than 7.
Landing on a number less than 8.
Landing on a number greater than 1.
None of the above.
RESULTS BOX:
3. What is the probability of choosing 14 hearts from a standard deck of 52 playing cards?
1
0
None of the above.
RESULTS BOX:
4. If a number is chosen at random from the following list, then what is the probability that it is prime?
2, 3, 5, 7, 11, 13, 17, 19
1
0
None of the above.
RESULTS BOX:
5. If a single 6-sided die is rolled, then which of the following events is neither certain nor impossible?
Rolling a number less than 7.
Rolling an even number.
Rolling a zero.
None of the above.
RESULTS BOX:
Experiment 1: What is the probability of each outcome when a dime is tossed?
Outcomes: The outcomes of this experiment are head and tail.Probabilities: P(head) = 12
P(tail) = 12
Definition: The sample space of an experiment is the set of all possible outcomes of that experiment.
The sample space of Experiment 1 is: {head, tail}
Experiment 2: A spinner has 4 equal sectors colored yellow, blue, green and red. What is the probability of landing on each color after spinning this spinner?
Sample Space: {yellow, blue, green, red}
Probabilities: P(yellow) = 14
P(blue) = 14
P(green) = 14
P(red) = 14
Experiment 3: What is the probability of each outcome when a single 6-sided die is rolled?
Experiment 4: A glass jar contains 1 red, 3 green, 2 blue and 4 yellow marbles. If a single marble is chosen at random from the jar, what is the probability of each outcome?
Sample Space: {red, green, blue, yellow}
Probabilities:
P(red) = 1
10
P(green) = 3
10
P(blue) = 2 = 110 5 P(yellow) = 4 = 210 5
Summary: The sample space of an experiment is the set of all possible outcomes for that experiment. You may have noticed that for each of the experiments above, the sum of the probabilities of each outcome is 1. This is no coincidence. The sum of the probabilities of the distinct outcomes within a sample space is 1.
The sample space for choosing a single card at random from a deck of 52 playing cards is shown below. There are 52 possible outcomes in this sample space.
The probability of each outcome of this experiment is: P(card) = 1 52
The sum of the probabilities of the distinct outcomes within this sample space is: 52 = 152
Exercises
Directions: Read each question below. Select your answer by clicking on its button. Feedback to your answer is provided in the RESULTS BOX. If you make a mistake, choose a different button.1. What is the sample space for choosing an odd
number from 1 to 11 at random?
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}
{1, 3, 5, 7, 9 11}
None of the above.
RESULTS BOX:
2. What is the sample space for choosing a prime
number less than 15 at random?
{2, 3, 5, 7, 11, 13, 15}
{2, 3, 5, 7, 11, 13}
{2, 3, 5, 7, 9, 11, 13}
All of the above.
RESULTS BOX:
3. What is the sample space for choosing 1 jelly bean at random from a jar containing 5 red, 7 blue and 2 green jelly beans?
{5, 7, 2}
{5 red, 7 blue, 2 green}
{red, blue, green}
None of the above.
RESULTS BOX:
4. What is the sample space for choosing 1 letter at random from 5 vowels?
{a, e, i, o, u}
{v, o, w, e, l}
{1, 2, 3, 4, 5}
None of the above.
RESULTS BOX:
5. What is the sample space for choosing 1 letter at
random from the word DIVIDE?
{d, i, v, i, d, e}
{1, 2, 3, 4, 5, 6}
{d, i, v, e}
None of the above.
RESULTS BOX:
Complementof an Event
Unit 6
Experiment 1: A spinner has 4 equal sectors colored yellow, blue, green and red. What is the probability of landing on a sector that is not red after spinning this spinner?
Sample Space: {yellow, blue, green, red}Probability: The probability of each outcome in this experiment is one
fourth. The probability of landing on a sector that is not red is the same as the probability of landing on all the other colors except red.P(not red) = 1 + 1 + 1 = 34 4 4 4
In Experiment 1, landing on a sector that is not red is the complement of landing on a sector that is red.
Definition: The complement of an event A is the set of all outcomes in the sample space that are not included in the outcomes of event A. The complement of event A is represented by (read as A bar).
Rule: Given the probability of an event, the probability of its complement can be found by subtracting the given probability from 1.P( ) = 1 - P(A)
You may be wondering how this rule came about. In the last lesson, we learned that the sum of the probabilities of the distinct outcomes within a sample space is 1. For example, the probability of each of the 4 outcomes in the sample space above is one fourth, yielding a sum of 1. Thus, the probability that an outcome does not occur is exactly 1 minus the probability that it does. Let's look at Experiment 1 again, using this subtraction principle.
Experiment 1: A spinner has 4 equal sectors colored yellow, blue, green and red. What is the probability of landing on a sector that is not red after spinning this spinner?
Experiment 2: A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a card that is not a king?
Probability: P(not king) = 1 - P(king)
= 1 - 4 52
=
4852
=
1213
Experiment 3: A single 6-sided die is rolled. What is the probability of rolling a number that is not 4?
Probability: P(not
4) = 1 - P(4)
= 1 - 16
= 56
Experiment 4: A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a card that is not a club?
Probability: P(not club) = 1 - P(club)
= 1 - 13
52
=
3952
= 34
Experiment 5: A glass jar contains 20 red marbles. If a marble is chosen at random from the jar, what is the probability that it is not red?
Probability: P(not red) = 1 - P(red)
= 1 - 1
= 0 Note: This is an impossible event.
Mutually Exclusive Events
Unit 6
Experiment 1: A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a 5 or a king?
Possibilities: 1. The card chosen can be a 5.2. The card chosen can be a king.
Experiment 2: A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a club or a king?
Possibilities: 1. The card chosen can be a club.2. The card chosen can be a king.3. The card chosen can be a king and a club (i.e., the
king of clubs).
In Experiment 1, the card chosen can be a five or a king, but not both at the same time. These events are mutually exclusive. In Experiment 2, the card chosen can be a club, or a king, or both at the same time. These events are not mutually exclusive.
Definition: Two events are mutually exclusive if they cannot occur at the same time (i.e., they have no outcomes in common).
Experiment 3: A single 6-sided die is rolled. What is the probability of rolling an odd number or an even number?
Possibilities: 1. The number rolled can be an odd number.
2. The number rolled can be an even number.
Events: These events are mutually exclusive since they cannot occur at the same time.
Experiment 4: A single 6-sided die is rolled. What is the probability of rolling a 5 or an odd number?
Possibilities: 1. The number rolled can be a 5.2. The number rolled can be an odd number (1, 3 or 5).3. The number rolled can be a 5 and odd.
Events: These events are not mutually exclusive since they can occur at the same time.
Experiment 5: A single letter is chosen at random from the word SCHOOL. What is the probability of choosing an S or an O?
Possibilities: 1. The letter chosen can be an S2. The letter chosen can be an O.
Events: These events are mutually exclusive since they cannot occur at the same time.
Experiment 6: A single letter is chosen at random from the word SCHOOL. What is the probability of choosing an O or a vowel?
Possibilities: 1. The letter chosen can be an O2. The letter chosen can be a vowel.3. The letter chosen can be an O and a
vowel.Events: These events are not mutually exclusive since they can occur at the same
time.
Summary: In this lesson, we have learned the difference between mutually exclusive and non-mutually exclusive events. We can use set theory and Venn Diagrams to illustrate this difference.
Mutually Exclusive Events Non-Mutually Exclusive EventsTwo events are mutually exclusive if they cannot occur at the same time (i.e., they have no outcomes in common).
Two events are non-mutually exclusive if they have one or more outcomes in common.
In the Venn Diagram above, the probabilities of events A and B are represented by two disjoint sets (i.e., they have no elements in common).
In the Venn Diagram above, the probabilities of events A and B are represented by two intersecting sets (i.e., they have some elements in common).
Addition Rulesfor Probability
Unit 6
Experiment 1: A single 6-sided die is rolled. What is the probability of rolling a 2 or a 5?
Possibilities: 1. The number rolled can be a 2.
2. The number rolled can be a 5.
Events: These events are mutually exclusive since they cannot occur at the same time.
Probabilities: How do we find the probabilities of these mutually exclusive events? We need a rule to guide us.
Addition Rule 1: When two events, A and B, are mutually exclusive, the probability that A or B will occur is the sum of the probability of each event.
P(A or B) = P(A) + P(B)
Let's use this addition rule to find the probability for Experiment 1.
Experiment 1: A single 6-sided die is rolled. What is the probability of rolling a 2 or a 5?
Probabilities: P(2) = 1
6
P(5) = 1
6
P(2 or 5) = P(2) + P(5)
= 1
+ 1
6 6
= 26
= 13
Experiment 2: A spinner has 4 equal sectors colored yellow, blue, green, and red. What is the probability of landing on red or blue after spinning this spinner?
Probabilities:
P(red) = 1
4
P(blue) = 1
4
P(red or blue) = P(red) + P(blue
)
= 1
+ 1
4 4
= 2
4
= 12
Experiment 3: A glass jar contains 1 red, 3 green, 2 blue, and 4 yellow marbles. If a single marble is chosen at random from the jar, what is the probability that it is yellow or green?
Probabilities:
P(yellow) = 4
10
P(green) = 3
10
P(yellow or green) = P(yellow
) + P(green)
= 4
+ 3
10 10
= 7 10
In each of the three experiments above, the events are mutually exclusive. Let's look at some experiments in which the events are non-mutually exclusive.
Experiment 4: A single card is chosen at random from a standard deck of 52 playing cards. What is the probability of choosing a king or a club?
Probabilities: P(king or club) = P(king) + P(club) - P(king of clubs)
= 4
+ 13
- 1
52 52 52
= 1652
= 4 13
In Experiment 4, the events are non-mutually exclusive. The addition causes the king of clubs to be counted twice, so its probability must be subtracted. When two events are non-mutually exclusive, a different addition rule must be used.
Addition Rule 2: When two events, A and B, are non-mutually exclusive, the probability that A or B will occur is:
P(A or B) = P(A) + P(B) - P(A and B)
In the rule above, P(A and B) refers to the overlap of the two events. Let's apply this rule to some other experiments.
Experiment 5: In a math class of 30 students, 17 are boys and 13 are girls. On a unit test, 4 boys and 5 girls made an A grade. If a student is chosen at random from the class, what is the probability of choosing a girl or an A student?
Probabilities: P(girl or A) = P(girl) + P(A) - P(girl and
A)
= 13
+ 9
- 5
30 30 30
= 1730
Experiment 6: On New Year's Eve, the probability of a person having a car accident is 0.09. The probability of a person driving while intoxicated is 0.32 and probability of a person having a car accident while intoxicated is 0.15. What is the probability of a person driving while intoxicated or having a car accident?
Probabilities: P(intoxicated or accident) = P(intoxicated) + P(accident) - P(intoxicated and accident)
= 0.32 + 0.09 - 0.15 = 0.26
Summary: To find the probability of event A or B, we must first determine whether the events are mutually exclusive or non-mutually exclusive. Then we can apply the appropriate Addition Rule:
Addition Rule 1: When two events, A and B, are mutually exclusive, the probability that A or B will occur is the sum of
the probability of each event.P(A or B) = P(A) + P(B)
Addition Rule 2:: When two events, A and B, are non-
mutually exclusive, there is some overlap between these events. The probability that A or B will occur is the sum of the probability of each event, minus the probability of the overlap.P(A or B) = P(A) + P(B) - P(A and B)
Independent Events
Unit 6
Experiment 1: A dresser drawer contains one pair of socks with each of the following colors: blue, brown, red, white and black. Each pair is folded together in a matching set. You reach into the sock drawer and choose a pair of socks without looking. You replace this pair and then choose another pair of socks. What is the probability that you will choose the red pair of socks both times?
There are a couple of things to note about this experiment. Choosing a pairs of socks from the drawer, replacing it, and then choosing a pair again from the same drawer is a compound event. Since the first pair was replaced, choosing a red pair on the first try has no effect on the probability of choosing a red pair on the second try. Therefore, these events are independent.
Definition: Two events, A and B, are independent if the fact that A occurs does not affect the probability of B occurring.
Some other examples of independent events are:
Landing on heads after tossing a coin AND rolling a 5 on a single 6-sided die. Choosing a marble from a jar AND landing on heads after tossing a coin. Choosing a 3 from a deck of cards, replacing it, AND then choosing an ace as the second
card. Rolling a 4 on a single 6-sided die, AND then rolling a 1 on a second roll of the die.
To find the probability of two independent events that occur in sequence, find the probability of each event occurring separately, and then multiply the probabilities. This multiplication rule is defined symbolically below. Note that multiplication is represented by AND.
Multiplication Rule 1: When two events, A and B, are independent, the probability of both occurring is:
P(A and B) = P(A) · P(B)
(Note: Another multiplication rule will be introduced in the next lesson.) Now we can apply this
rule to find the probability for Experiment 1.
Experiment 1: A dresser drawer contains one pair of socks with each of the following colors: blue, brown, red, white and black. Each pair is folded together in a matching set. You reach into the sock drawer and choose a pair of socks without looking. You replace this pair and then choose another pair of socks. What is the probability that you will choose the red pair of socks both times?
Probabilities:
P(red) = 1
5
P(red and red) = P(red) · P(red)
= 1
· 1
5 5
= 1 25
Experiment 2: A coin is tossed and a single 6-sided die is rolled. Find the probability of landing on the head side of the coin and rolling a 3 on the die.
Probabilities:
P(head) = 1
2
P(3) = 1
6
P(head and 3) = P(head
) · P(3)
= 1 · 1
2 6
= 1 12
Experiment 3: A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a jack and an eight?
Probabilities:
P(jack) = 4
52
P(8) = 4
52
P(jack and 8) = P(jack) · P(8)
= 4
· 4
52 52
= 16 2704
= 1 169
Experiment 4: A jar contains 3 red, 5 green, 2 blue and 6 yellow marbles. A marble is chosen at random from the jar. After replacing it, a second marble is chosen. What is the probability of choosing a green and a yellow marble?
Probabilities:
P(green) = 5
16
P(yellow) = 6
16
P(green and yellow) = P(green) · P(yellow)
= 5
· 6
16 16
= 30 256
= 15 128
Each of the experiments above involved two independent events that occurred in sequence. In some cases, there was replacement of the first item before choosing the second item; this replacement was needed in order to make the two events independent. Multiplication Rule 1 can be extended to work for three or more independent events that occur in sequence. This is demonstrated in Experiment 5 below.
Experiment 5:
A school survey found that 9 out of 10 students like pizza. If three students are chosen at random with replacement, what is the probability that all three students like pizza?
Probabilities:
P(student 1 likes pizza) = 9
10
P(student 2 likes pizza) = 9
10
P(student 3 likes pizza) = 9
10
P(student 1 and student 2 and student 3 like pizza) =
9 ·
9 ·
9 =
729 10
10
10
100
All of the experiments above involved independent events with a small population (e.g. A 6-sided die, a 2-sided coin, a deck of 52 cards). When a small number of items are selected from
a large population without replacement, the probability of each event changes so slightly that the amount of change is negligible. This is illustrated in the following problem.
Problem: A nationwide survey found that 72% of people in the United States like pizza. If 3 people are selected at random, what is the probability that all three like pizza?
Solution: Let L represent the event of randomly choosing a person who likes pizza from the U.S.P(L) · P(L) · P(L) = (0.72)(0.72)(0.72) = 0.37 = 37%
In the next lesson, we will address how to handle non-replacement in a small population.
Dependent Events Unit 6
Experiment 1: A card is chosen at random from a standard deck of 52 playing cards. Without replacing it, a second card is chosen. What is the probability that the first card chosen is a queen and the second card chosen is a jack?
Analysis: The probability that the first card is a queen is 4 out of 52. However, if the first card is not replaced, then the second card is chosen from only 51 cards. Accordingly, the probability that the second card is a jack given that the first card is a queen is 4 out of 51.
Conclusion: The outcome of choosing the first card has affected the outcome of choosing the second card, making these events dependent.
Definition: Two events are dependent if the outcome or occurrence of the first affects the outcome or occurrence of the second so that the probability is changed.
Now that we have accounted for the fact that there is no replacement, we can find the probability of the dependent events in Experiment 1 by multiplying the probabilities of each event.
Experiment 1:
A card is chosen at random from a standard deck of 52 playing cards. Without replacing it, a second card is chosen. What is the probability that the first card chosen is a queen and the second card chosen is a jack?
Probabilities:
P(queen on first pick) = 4
52
P(jack on 2nd pick given queen on 1st pick)
= 4
51
P(queen and jack) = 4
· 4
= 16
=
52
51
2652
Experiment 1 involved two compound, dependent events. The probability of choosing a jack on the second pick given that a queen was chosen on the first pick is called a conditional probability.
Definition: The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already occurred. The notation for conditional probability is P(B|A) [pronounced as The probability of event B given A].
The notation used above does not mean that B is divided by A. It means the probability of event B given that event A has already occurred. To find the probability of the two dependent events, we use a modified version of Multiplication Rule 1, which was presented in the last lesson.
Multiplication Rule 2: When two events, A and B, are dependent, the probability of both occurring is:
P(A and B) = P(A) · P(B|A)
Let's look at some experiments in which we can apply this rule.
Experiment 2: Mr. Parietti needs two students to help him with a science demonstration for his class of 18 girls and 12 boys. He randomly chooses one student who comes to the front of the room. He then chooses a second student from those still seated. What is the probability that both students chosen are girls?
Probabilities: P(Girl 1 and Girl 2) = P(Girl 1) and P(Girl 2|Girl 1)
= 18
· 17
30 29
= 306870
= 51 145
Experiment 3: In a shipment of 20 computers, 3 are defective. Three computers are randomly selected and tested. What is the probability that all three are defective if the first and second ones are not replaced after being tested?
Experiment 4: Four cards are chosen at random from a deck of 52 cards without replacement. What is the probability of choosing a ten, a nine, an eight and a seven in order?
Probabilities: P(10 and 9 and 8 and 7) = 4
· 4
· 4
· 4
= 256
= 32
52 51 50 49 6,497,400 812,175
Experiment 5: Three cards are chosen at random from a deck of 52 cards without replacement. What is the probability of choosing 3 aces?
Probabilities: P(3 aces) = 4
· 3
· 2
= 24
= 1
52 51 50 132,600 5,525
Conditional Probability
Unit 6
Problem: A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed the second test?
Analysis: This problem describes a conditional probability since it asks us to find the probability that the second test was passed given that the first test was passed. In the last lesson, the notation for conditional probability was used in the statement of Multiplication Rule 2.
Multiplication Rule 2: When two events, A and B, are dependent, the probability of both occurring is:
The formula for the Conditional Probability of an event can be derived from Multiplication Rule 2 as follows:
Start with Multiplication Rule 2.
Divide both sides of equation by P(A).
Cancel P(A)s on right-hand side of equation.
Commute the equation.
We have derived the formula for conditional probability.
Now we can use this formula to solve the problem at the top of the page.
Problem: A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed the second test?
Solution: P(Second|First) =
P(First and Second) =
0.25 = 0.60 = 60%P(First) 0.42
Let's look at some other problems in which we are asked to find a conditional probability.
Example 1: A jar contains black and white marbles. Two marbles are chosen without replacement. The probability of selecting a black marble and then a white marble is 0.34, and the probability of selecting a black marble on the first draw is 0.47. What is the probability of selecting a white marble on the second draw, given that the first marble drawn was black?
Solution: P(White|Black) =
P(Black and White) = 0.34 = 0.7
2 = 72%P(Black) 0.47
Example 2: The probability that it is Friday and that a student is absent is 0.03. Since there are 5 school days in a week, the probability that it is Friday is 0.2. What is the probability that a student is absent given that today is Friday?
Example 3: At Kennedy Middle School, the probability that a student takes Technology and Spanish is 0.087. The probability that a student takes Technology is 0.68. What is the probability that a student takes Spanish given that the student is taking Technology?
Solution: P(Spanish|Technology) =
P(Technology and Spanish) = 0.087 = 0.13 = 13%
P(Technology) 0.68
Summary: The conditional probability of an event B in relationship to an event A is the probability that event B occurs given that event A has already occurred. The notation for conditional probability is P(B|A), read as the probability of B given A. The formula for conditional probability is:
The Venn Diagram below illustrates P(A), P(B), and P(A and B). What two sections would have to be divided to find P(B|A)? Answer
Question 1. Two cards are drawn from a pack of 52 cards in succession. Find the probability that both are kings when (i) The first drawn card is replaced (ii) The card is not replacedSolution: Let A be the event of drawing a king in the first draw. B be the event of drawing a king in the second draw. (i) Card is replaced: n(A) = 4 (king) n(B) = 4 (king) and n(S) = 52 (Total) Clearly the event A will not affect the probability of the occurrence of event B and therefore A and B are independent. P(A ∩ B) = P(A) * P(B) = 4/52 × 4/52 P(A ∩ B) =1/169 (ii) (Card is not replaced)
In the first draw, there are 4 kings and 52 cards in total. Since the king, drawn at the first draw is not replaced, in the second draw there are only 3 kings and 51 cards in total. Therefore the first event A affects the probability of the occurrence of the second event B.
Therefore A and B are not independent they are dependent events. Therefore P (A ∩ B) = P(A) * P(B/A) P (A) = 4/52; P (B/A) = 3/51 P (A ∩ B) = P(A) * P (B/A) = 4/52 x 3/51
P(A ∩ B) = 1/221Question 2. The total weight of 8 people chosen at random follows a normal distribution with a mean of 550kg and a standard deviation of 150kg. What’s the probability that the total weight of 8 people exceeds 600kg?Solution: Here X = 600kg m, the mean = 500kg s, the standard deviation = 150kg. Z = (600 – 550)/150 = 50/150 Z = 0.33Z scores from normal distribution table is 0.3707This is the probability that the weight will exceed 600kg.
Answer: The probability that the total weight of 8 people exceeds 600kg is 0.37.
Example 1: A bag contains 25 bulbs, out of 25 bulbs, 8 bulbs are defective, 4 bulbs are chosen at random from this bag. Find the probability that at least two of these is defective.Solution:Out of 25 bulbs, 8 bulbs are defective.17 bulbs are favorable bulbs.E = event for getting no bulb is defective.n(E) = 17C4
Out of 17 bulbs we have to choose 4 bulbs randomly, so the number of ways = 17C4.n(E) = 17C4
n(S) = 25C4
P(E) = 17C4/25C4
= 238/1265Probability of at least two is defective + probability of two is non defective = 2P(E) + p(E) = 2238/1265 + p(E) = 2P(E) = 2292/1265
Example 2: A bag contains 8 red and 6 orange bulbs. 2 bulbs are drawn at random. Find the probability that they are of the same color.Solution:Let S be the sample spaceNumber of ways for drawing 2 bulbs out of 8 red and 6 orange bulbs = 14C2
= 91n(S) = 91Let E = event of getting both bulbs of the same colorThenn(E) = number of ways of drawing (2 bulbs out of 8) or (2 balls out of 6)= 8C2 + 6C2
= 28 + 15 = 43P(E) = n(E)/n(S) = 43/91
Practice QuestionsSolve these practice questions. These questions are very easy to solve.Question 1: Find the probability of drawing 4 bulbs out of 6 bulbs in bag.Answer: 2/3Question 2: Find the probability of drawing 2 bulbs out of 4 red bulbs and 4 orange bulbs in a bag.Answer: 1/4.
Q 1: A coin is tossed. If it shows a tail we draw a ball from a box which contains red and 3 black balls ; if ti shows a head, we throw a die.Sol : Filet red balls named as R1 and R2, and the black balls be named as B1, B2, B3.Then the sample space is S={TR1, Tr2, TB1,TB2,TB3,H1,H2,H3,H4,H5,H6}Q 2: Three unbiased coins are tossed once. What is that probability of gettingi) All heads?ii) Two heads?iii) One head?iv) At least 1 head?Sol : In tossing three coins, the sample space is given byS= {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}And therefore n(S)=8.(i) Getting all headLet E1 = event of getting all heads. Then,E1= {HHH} and, therefore, n (E1) =1.Therefore P (getting all heads) = P(E1) = n(E1) / n(S) = 1 / 8.(ii) Getting 2 headLet E2 = event of getting 2 heads. Then,E2= {HHT, HTH, THH} and, therefore, n (E2) =3.Therefore P (getting 2 heads) = P(E2) = n(E2) / n(S) = 3 / 8.(iii) Getting one headLet E3 = event of getting one head. Then,E3= {HTT, THT, TTH} and, therefore, n (E3) =3.
Therefore P (getting one head) = P (E3) = n(E3) / n(S) = 3 / 4(iv) Getting of at least 1 headLet E4 = event of getting at least one head. Then,E4= {HTT, THT, TTH, HHT, HTH, THH, HHH} and, therefore, n (E4) =7.Therefore P (getting at least one head) = P (E4) = n(E4) / n(S) = 7 / 8.Q 1:If P (A) =7 P (B) = 2/15 and P (A ∩B ) = 4/15, evaluate P(A|B).Sol:From conditional probability definition we know that, P (A|B) = P (A ∩B) P (B)Hence,
P(A|B) =
= * = = .Q 2: If P(A|B) = 6 , P(B)= 8 , Find P (A ∩B ) .Sol:From conditional probability definition we know that, P (A|B) = P (A ∩B) P (B)
6 = P (A ∩B ) = 8 x 6 = 48
Word Problems on Solving Conditional Probability QuestionsSolving word problems conditional probability questions is simple , first we convert them to the simplest form , then plug in the values .
Q 1: Ten cards numbered 1 to 10 are placed in a deck, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number?Solve using conditional probability.Sol:Let A be the event ‘the number on the drawn card is even’ and B be the event ‘the number on the card drawn is greater than 3’. We have to solve P(A|B).Now, the sample space of the experiment is S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}Then A = {2, 4, 6, 8, 10}, B = {4, 5, 6, 7, 8, 9, 10} and A ∩ B = {4, 6, 8, 10}
Also P (A) = , P (B) = and P (A ∩ B) = From conditional probability definition we know that, P (A|B) = P (A ∩B) P (B)
P (A|B) = = .Q 2: In an industry, there are 1000 employees, out of which 430 are women. It is known that out of 430, 10% of the women works in finance section.. What is the
probability that an employee chosen randomly works in finance section given that the chosen employee is a woman?Solve using conditional probability.Sol: Let A denote the event that an employee chosen randomly is work is finance section, and B be the event that the randomly chosen employee is woman.Now we have to solve P (A|B)
We know that , P(B) = = 0.43
P(A ∩ B) = { 10% of 430} = 0.043Then , P (A|B) = P (A ∩B) {from conditional probability definition} P (B) = 0.043/ 0.43 = 0.1P(A|B) = 0.1Study about Probability QuestionsExperiment 1: Tossing a coinPossible outcomes are head or tail.Sample space, S = {head, tail}.Experiment 2: Tossing a diePossible outcomes are the numbers 1, 2, 3, 4, 5, and 6Sample space, S = {1, 2, 3, 4, 5, 6}
Study some Probability Example QuestionsQuestion 1: Two players, John and Jim, play a tennis match. It is known that the probability of John winning the match is 0.62. What is the probability of Jim winning the match?Solution:In this question,We can assume S and R as the events that John wins the match and Jim wins the match, respectively. The probability of John’s winning = P(S) = 0.62 (given) The probability of Jim’s winning = P(R) = 1 – P(S) [As the events R and S are complementary] = 1 – 0.62 = 0.38Question 2: A box contains 3 blue, 2 white, and 4 red marbles. If a marble is strained at chance from the box, what is the probability that it will be (i) White? (ii) Blue? (iii) Red?Solution: The question says that a marble is drawn at chance is a short way of saying that all the marbles are equally likely to be strained Therefore, the number of possible outcomes = 3 +2 + 4 = 9 Let W represent the event ‘the marble is white’, B denote the experience ‘the marble is blue’ and R denote the event ‘marble is red’. (i) The number of outcomes favorable to the event W = 2 So, P (W) = 2 / 9Similarly, (ii) P (B) = 3 / 9
= 1 / 3 And,(iii) P(R) = 4 / 9Note that: P (W) + P (B) + P (R) = 1.
Probability Sample Question for Preparation of Exam:Here are few examples for probabilty preparation,Question 1 for Preparation: In a set of 6 numbers, (1) Find the probability of an even number (2) Find the probability of an odd number. Exam the probability of each of these events. Solution: Probability P( E ) =
P( E1 ) = = 0.5
P( E2 ) = = 0.5 Question 2 for preparation: There are 200 values in the Diary. The frequency distribution of 8 at their unit place digit. Numbers 1 2 3 4 5 6 7 8 9 0 Frequency 23 27 21 20 22 9 15 27 15 21 The number is chosen at random. Exam what is the probability getting the unit place is 8? Solution: The probability of digit 8 is the unit place. Probability P( E ) =
= P(E) = 0.135
Probability Exercises Questions for Preparation of Exam:1). A rubber company kept a record of covering distance before when tyre needed to be replaced. Given data shows the results of 1000 cases. unit more than 13000 8001 to 13000 3000 to 8000 less than 3000 Freq 445 325 210 20 If you buy a tyre of this company, calculate the probability that : (i) replaced before it has covered 3000? (ii) more than 8000? Answer: (i) covered 3000 (km) P(E) = 0.02 (ii) more than 8000 (km) = 0.772). The percentage of marks of a student in the unit tests are Unit test I II III IV V Percentage 77 52 76 90 56 Exam what is the probability of the student gets more than 70% marks in a unit test.
Answer : more than 70% marks P(E) = 0.Each and very letter MADAM is written on a card. A card is select at random from the container. Find the probability of getting the letter M?Solution:The card is selected randomly; it means that each card has the same probability of selected.Sample space S = {M, A, D, A, M} = 5There are two card have the letter MLet E= event of getting the letter M = {M1, M2} = 2Probability formula = Number of favorable outcomes of an event Number of total outcomesNumber of favorable outcomes of an event = 2Number of total outcomes =5
P (E) = Question:Each and very letter MADAM is written on a card. A card is select at random from the container. Find the probability of getting the letter D?Solution:The card is selected randomly; it means that each card has the same probability of selected.Sample space S = {M, A, D, A, M} = 5There are one card have the letter DLet E= event of getting the letter D = {D} = 1Probability formula = Number of favorable outcomes of an event Number of total outcomesNumber of favorable outcomes of an event = 1Number of total outcomes =5
P (E) =
5th Grade Math Probability Example ProblemsQuestion:Each and very letter HELLO is written on a card. A card is select at random from the container. Find the probability of getting the letter L?Answer:
Question:Each and very letter HELLO is written on a card. A card is select at random from the container. Find the probability of getting the letter O?Answer:
n a bakery, there are 50 biscuits, 20 bread packets and 70 cakes. What is the probability of buying biscuit or bread?Solution:Given:Number of biscuits = 50Number of bread packets = 20Number of cakes = 70Step 1:Total = 50 + 20 + 70n(S) = 140Step 2:P(A) – buying Biscuits
P(A) =
=
= Step 3:P(B) – buying bread packets
P(B) =
=
= Step 4: P(buying biscuit or bread)P(A B) = P(A) + P(B)
= +
= = 2Answern a Jewellery shop, there are 50 bangles, 40 studs and 60 rings. Write the probability of the following options.a) P (Not selecting a ring)b) P (Selecting a bangle or stud)Solution:Given:Number of bangles = 50Number of studs = 40Number of rings = 60Step 1:
= b) P(bangle or stud)P(A or B) = P(AUB)= P(A) + P(B)
= +
=
=
Practice Problems to Preparation for Probability Rules Exam:Problem: 1
P(C) = . Find P(C')
Answer: P(A') = Problem: 2
P(A) = , P(B) = and A and B are independent events. Find P(B') , P(A B).
Answer: P(A') = P(B') = In a shop there are 7 bikes and 3 cars. If one vehicle is selected at random and then a second vehicle is selected at random, what are the chances that both of vehicles will be bike?Solution: Total number of bikes = 7 Total number of cars = 3 Total number of vehicles n(s) = 7 + 3 = 10 For the first time, there are 7 chances in 10 vehicles
So probability for the first chance P1 = For the second time, there are 6 chances in 9 vehicles
Probability for the second chance P2= Probability for both chance to get a bike = P1 * P2
=
=
=
Two coins are tossed once. Find the probability of (i) getting two tails (ii) getting at least one tail (iii) getting no tail (iv) getting one tail and one headSolution: Let 'H' denote the event of getting a head and 'T' denote the event of getting a tail. On tossing two coins simultaneously, all possible outcomes are HH, HT, TH, and TT.Numbers of possible outcomes = 4 (i) Favorable outcomes of two tails is (T, T) Number of favorable outcomes = 1
P (Two tails) = (ii) Favorable outcomes of at least one tail are HT, TH, and TT.
Number of favorable outcomes = 3
P (at least one tail) = (iii) Getting no tail is (H, H). Favorable outcome is 1.
P (no tail) = (iv) Favorable outcomes of at least one tail and one head HT, TH, Number of favorable outcomes = 2
P (at least one tail and one head) = --- >The shop has 82 products. In those products, there are 26 car toys, 29 train toys and 27 bus toys are available. What is the probability for the following outcomes? i) Select the car toys. ii) Select the train toys.Solution: Number of products n(S) = 82 Number of car toys n(A) = 26 Number of train toys n(B) = 29 Number of bus toys n(C) = 27 i) Assume the P(A) is the probability for select the car toys.
P(A)=
=
= . ii) Assume the P(B) is the probability for select the train toys.
P(B) =
= . In the flower garden, there are 52 flowers are available. In those flowers, 10 are the red roses, 15 are the white roses and the remaining are the yellow roses. i) Choose the red roses ii) Choose the white rosesSolution: Number of roses n (S) = 52 Number of red roses n(A) = 10 Number of white roses n(B) = 15 We don’t know the yellow roses quantity. So find that one with the help of other quantity. Number of yellow roses n(C ) = 52 - (10 + 15) = 52 - 25 n(C) = 27. i) Assume P(A) is the probability for choose the red roses.
P(A) =
=
= . ii)Assume P(B) is the probability for choose the white roses.
P(B) =
= .
Practice Problems - Probability Test Prep Tutoring:Test question 1Julie has the 10 pens, Jessie has the 12 pens. What is the probability for select the Jessie’s pens?Answer:
Probability =
= .Test question 2 Alex has one bag. In that bag, there are 15 red balls and 22 white balls are available. What is the probability for select the red balls?Answer:
Probability = .Test question 3: The shop has 12 fruits. In those fruits, there are 4 apples and 8 mangos. What is the probability for select the mangos?Answer:
Probability = .
= .Solved problem: Pierre throw a die once. Try to find out the probability of getting a number less than or equal to three. And also find the probability of getting a number greater than three.Solution: Pierre throws a die once. Let us take ‘probability of getting a number greater than three’ is G, and ‘the probability of getting a number less than or equal to three’ is L. Number of possible outcomes = 6
P(G) = P(number greater than 3) = = Outcomes favorable to the event L are 1, 2, 3. So, the number of outcomes favorable to L is 3.
Therefore, P(L) = =Answer: The probability of getting a number less than or equal to three and the
probability of getting a number greater than three is and respectively.
Exam Questions - Prepare for Ks-4 Probability Exam:Exam question 1: Try to determine the probability of getting a head when a coin is tossed once.Exam question 2: In a shuffle deck of 52 cards, Patrick took one card without look it. Calculate the probability that he took an ace.Exam question 3: Two players, Imam and Raman, play a chess game. The probability of Imam winning the match is 0.4. Calculate the probability of Raman winning the match.Exam question 4: A die is rolling by Minnie. Let A be the event ‘getting a prime number’, B be the event ‘getting an odd number’. Calculate the sets representing the events A or B.Exam question 5: An urn contains 10 black and 5 white balls. Without replacing two balls are drawn from the urn. Find the probability that both drawn balls are black.Answer keys - prepare for ks-4 probability exam:
Answer key 1: The probability of getting a head when a con is tossed is .
Answer key 2: The probability that Patrick took an ace is Answer key 3: The probability of Raman winning the match is 0.6.Answer key 4: ‘A or B’ = A U B = {1, 2, 3, 5}
Answer key 5: In a bags, there are filled with 10 red rose flowers, 15 yellow rose flowers and 25 pink rose flowers, Find the probability of the red, yellow and pink color flowers?Solution:From the given data’s we can find the total number of flowers.So total number of flowers =number of red color flower + yellow color flower + pink color flowerTotal number of red color flowers = 10Total number of yellow color flowers = 15Total number of pink color flowers = 25So the total number of red, yellow and pink color flowers is 50Probability of taking red color flowers = 10/50 = 1/5Probability of taking yellow color flowers = 15/50 = 3/10Probability of taking pink color flowers = 25/50 = 1/2Example 2 – math help with probability for elementary school:If the die (single) will be rotate find the Probability of receiving each number?Solution:The die has 6 sides it will be number as 1, 2,3,4,5 and 6 like that
The probability for the occurrence of each an every number are shown below:Probability of receiving number 1 is P (1) =1/6Probability of receiving number 2 is P (2) =1/6Probability of receiving number 3 is P (3) =1/6Probability of receiving number 4 is P (4) =1/6Probability of receiving number 5 is P (5) =1/6Probability of receiving number 6 is P (6) =1/6(Since single die has only one 1, 2,3,4,5 and 6 so the probabilities are the same)Example 3 – math help with probability for elementary school:Suppose the rectangle is divided into 15 equal parts and those parts are colored using different colors, 5 red, 7blue, 3 yellow?Solution:Here the given problem is called as experiment, and outcome of those experiment are pink, yellow and blueNow we can find the probabilities of each colorP (red) =5/15P (blue) =7/15P (yellow) =3/15e available. What is the probability for the following outcomes? i) Select the dolls. ii) Select the hats.Solution: Number of products n(S) = 72 Number of dolls n(A) = 20 Number of hats n(B) = 31 Number of boxes n(C) = 21 i) Assume the P(A) is the probability for select the dolls. P(A) = `(n(A))/(n(S))` = `(20)/(72)` = `(5)/(18)` . ii) Assume the P(B) is the probability for select the hats. P(B) =`(n(B))/(n(S))` = `(31)/(72)` .Example 2: The dress shop has the 45 dresses. In those dresses there are 20 gens dresses, 12 are the ladies dresses and the remaining are the kids dresses. What is the probability for the following outcomes? i) Choose the gens dresses ii) Choose the ladies dressesSolution: Number of dresses n (S) = 45 Number of gens dresses n(A) = 20
Number of ladies dresses n(B) = 12 We don’t know the kids dress quantity. So find that one with the help of other quantity. Number of kids dresses n(C ) = 45 - (20+12) = 45 - 32 n(C) = 13. i) Assume P(A) is the probability for choose the gens dresses. P(A) =`(n(A))/(n(S))` =`(20)/(45)` =`(4)/(9)` . ii)Assume P(B) is the probability for choose the ladies dresses. P(B) =`(n(B))/(n(S))` = `(12)/(45)` = `(4)/(15)` .
Test Problems - Online Probability Test Prep Tutor:Test question 1: Stephen has a 17 balls and John has the 12 balls. What is the probability for select the Stephen’s balls ?Answer: Probability = `(17)/(29)` .Test question 2: Michel has 23 apples and Jenifer has the 26 apples. What is the probability for select the Michel’s apples? Answer: Probability = `(23)/(49)` .Test question 3: The shop has 16 rings. In those rings, there are 7 diamond rings and 9 gold rings. What is the probability for select the gold rings?Answer: Probability = `(9)/(16)` .Consider A be the Poisson’s lambda (2) and B which has the Poisson’ lambda (3) and let C = A + B. Calculate the distribution of A conditional on C = 10.Solution:A: lambda = 2B: lambda = 3The value of C = A + B = 3 + 2 = 5Using the formula of conditional probability Poisson distribution
= =
=
= = 0.04031Answer: 0.04031Example problem 2 – Conditional probability Poisson learningConsider X be the Poisson’s lambda (3) and Y which has the Poisson’ lambda (2) and let Z = X - Y. Calculate the distribution of X conditional on Z = 5.Solution:X: lambda = 3Y: lambda = 2The value of Z = X - Y = 3 – 2 = 1Using the formula of conditional probability Poisson distribution
= = = = 4.3946Answer: 4.3946
Practicing Problems – Conditional Probability Poisson LearningPracticing problem 1 – Conditional probability poisson learningConsider P be the Poisson’s lambda (2) and Q which has the Poisson’ lambda (2) and let R = P/ Q. Calculate the distribution of P conditional on R = 2.Answer: 0.398Practicing problem 2 – Conditional probability poisson learningConsider X be the Poisson’s lambda (3) and Y which has the Poisson’ lambda (2) and let Z = X* Y. Calculate the distribution of X conditional on Z = 3.Answer: 4.3946Example problem 1- learn conditional probability distribution
Problem 1: If A and B be two events, P(B) = and P(A and B) = Solution:Step 1: Given:
A and B = Events
P( A and B ) =
P( B ) = Step 2: To find:P( B | A )Step 3: Formula:
Conditional Probability formula = P(B | A) = Step 3: Solve:
P( B | A ) =
=
=
Result: Conditional Probability distribution = Example Problem 2- learn conditional probability distribution
If A and B be two events, P(B) = and P(A and B) = Solution:Step 1: Given:
A and B = Events
P( A and B ) =
P( B ) = Step 2: To find:P( B | A )Step 3: Formula:Conditional Probability formula = P(B | A) = Step 3: Solve:
P( B | A ) =
=
=
=
Result: Conditional Probability distribution =
Practice Problems to Learn Conditional Probability Distribution:Problem 1- learn conditional probability distribution
If A and B be two events, P(B) = and P(A and B) = . Then find the value of P(B /A)?
Problem 2- learn conditional probability distribution
If A and B be two events, P(B) = and P(A and B) = Then find the value of P(B /A)?
r: 1/5
Example to Learn Conditional Probability Proof:In the population of England, the probability the Cricket player’s life is at least 85 years is 0.85 and 0.75 if the player won’t live more than 95 years. If a player has 85 years old, solve the simple conditional probability that he will survive on 95 years. If A subset of B then, P (X and Y) = P (X)Solution: Let us take X is the event that the player lives to 80 years and Y is the event that the player will live at least 80 years. So given that P ( Y ) = 0.75 and P ( X ) = 0.85 So Conditional probability,P ( Y | X ) = P ( X and Y ) / P ( X ) The given condition is P ( X and Y ) = P ( Y ) = 0.75 P ( X | Y ) = 0.75 / 0.85 P ( X | Y ) = 0.88Example Problems to Learn Conditional Probability Rule:Example: 1
Find the value of P(B |A) where P(A) = and P(A and B) = .Solution:Step 1:
P( A and B ) =
P(A) = Step 2: To find:P( B | A )Step 3: Formula:
Conditional Probability = P(B | A) = Step 3: Solve:
P( B | A ) =
=
=
=
Answer: Conditional probability =
Example: 2
Find the value of P(B |A) where P(A) = and P(A and B) = . Solution:Step 1:
P( A and B ) =
P( A ) = Step 2: To find:P( B | A )Step 3: Formula:
Conditional Probability = P(B | A) = Step 3: Solve:
P( B | A ) =
=
=
=
Answer: Conditional Probability = Example: 3
Find the value of P(B |A) where P(A) = and P(A and B) = . Solution:Step 1:
P( A and B ) =
P(A) = Step 2: To find:P( B | A )Step 3: Formula:
Conditional Probability = P(B | A) = Step 3: Solve:
P( B | A ) =
=
=
Answer: Conditional probability =
Practice Problems to Learn Conditional Probability Rule:Problem: 1
Find the value of P(B |A) where P(A) = and P(A and B) = .
Answer: Problem: 2
Find the value of P(B |A) where P(A) = and P(A and B) = .Answer: 5
Events of Conditional Probability: For example, we previously calculated the probability of rolling a 5 above. Now say we want to work out the probability of rolling a 5 given that one or both of the dice rolled is a 2. We would calculate this conditional probability like so A = {(1, 4), (2, 3), (3, 2), (4, 1)} B = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (1, 2), (3, 2), (4, 2), (5, 2), (6, 2)} A AND B = {(2, 3), (3, 2)} P(A | B) = P(A AND B) / P(B) = P({(2, 3), (3, 2)}) / P(B) = (1/18) / (11 / 36) = (2/11)
Examples for Conditional Probability:Example 1: A math teacher gave her class two tests. 25% of the class passed both tests and 42% of the class passed the first test. What percent of those who passed the first test also passed he second test? Solution: P(Second | First) = P(First and Second) / P(First) = 0.25/0.42 = 0.60 = 60%Example 2: A jar contains black and white marbles. Two marbles are chosen without replacement. The probability of taking a black marble and then a white marble is 0.34, and the probability of taking a black marble on the first draw is 0.47. What is the probability of taking a white marble on the second draw, given that the first marble drawn was black? Solution: P(White | Black) = P(Black and White) / P(Black) = 0.34 / 0.47 = 0.72 = 72%Example 3: At The Middle School, the conditional probability that a student selects Technology and Spanish is 0.087. The conditional probability that a student takes Technology is 0.68. What is the probability that a student selects Spanish given that the student is taking Technology?Solution:
uestion 30Question: A card is drawn from a well-shuffled pack of cards and its suit is noted. (spade, clubs, hearts or diamond). This trial is repeated 400 times and the Number of times each suit is drawn is given below.
When a card is drawn at random what is the probability that it is1) a diamond2) a black card3) not a spadeAnswer: 1) Total number of trials = 400Number of trials in which a diamond showed up = 76
P (the card drawn is a diamond)
2) Number of times a black card showed up = 96 + 108 = 204
P (the card drawn is black)
3) Number of times a card other than a spade showed up = 108 + 120 + 76 = 304
P (the card is not a spade)
Question 31Question: The monthly wages of 200 workers in a factory is given by the following table:
If one worker in the factory is chosen at random, what is the probability that1) his wage is in the range Rs. 3500 - 4500?
2) his wage is Rs. 4000 or above?3) Give two events in the content, one having probability 0 and the other having probability 1.
Answer: Total number of workers = 200
1) Number of workers whose wage is between Rs. 3500 and Rs. 4500
= 63 + 42 = 105
P (the worker's wage is between Rs. 3500 and Rs. 4500)
2) Number of workers with wage above Rs. 4000
= 42 + 19 = 61
P (the worker's wage is above Rs. 4000)
3) Let E1 be the event "the wage of the worker is less than Rs. 3000".
Since there is no worker whose wage is less than Rs. 3000, P(E1) = 0.
Let E2 be the event: "the wage is not less than Rs. 3000".
As the wages of all the workers considered is greater than or equal to Rs. 3000, P(E2) = 1
Question 32Question: In a cricket match, the number of dot-balls bowled in an over was noted down in the first 30 overs as follows:
What is the probability that an over bowled in the match?1) has no dot balls2) has not more than 2 dot balls3) at least two dot balls?
Answer: Total number of overs = 30
1) Number of overs with no dot balls = 7
P (the over has no dot balls)
2) Number of overs with not more than 2 dot balls = 7 + 5+ 5 = 17
P (the over has not more than 2 dot balls)
3) Number of overs with atleast 2 dot balls = 5 + 4 + 3 + 3 + 3 = 18
P (the over has atleast 2 dot balls)
Question 33Question: 100 mangoes are selected at random from each of 5 baskets of mangoes, and the number of mangoes which are spoilt is counted and recorded as follows:
When one such basket is checked what is the probability that it has1) no spoilt mangoes2) Atleast 10 spoilt mangoes3) more than 20 spoilt mangoes
Answer: Number of baskets = 5
1) Number of baskets with no spoilt mangoes = 0
P (the basket has no spoilt mangoes) = 0
2) No of basket has no spoilt mangoes = 5
P (the basket has at least 10 spoilt mangoes) =1
3) No of baskets with more than 20 spoilt mangoes = 3
P(the basket has more than 20 spoilt mangoes)
Question 34Question: Consider the following frequency distribution table which gives the heights of 40 students in a class
Find the probability that the height of a student in the class1) lies in the interval 150 - 1552) is 145 cm or above 145 cm3) is below 150 cm
Answer: Total Number of students = 40
1) Number of students whose height lies in the interval 150 - 155 cm = 12
P (the height of the student is in the interval 150 - 155)
2) Number of students whose height is 145 cm or above 145 cm
= 11 +12 + 9 = 32
P (the height of the student is 145 cm or above)
3) Number of students whose height is below 150 cm= 8 + 11 = 19
P (the height of the student is less than 150 cm)
Question 35Question: Cards numbered 1 to 10 are placed in a box. One card is drawn and the number noted. This trial is repeated 500 times and the result is tabulated as follows:
If a card is drawn what is the probability that the card no is1) a prime number2) an even number3) not less than 8
Answer: Number of trials = 500
1) Number of trials in which the card drawn has a prime number
= 34 + 52 + 54 + 60 = 200
P(the card number is prime)
2)Number of trials in which the card drawn has an even number
= 34 + 48 + 36 + 62 + 70 = 250
P(the card number is even)
3)Number of trials in which the card number is greater than 6
= 60 + 62 + 56 + 70 = 248
P (card number is greater than 6)
4) Number of trials in which the card number is not less than 8
= 62 + 56 + 70 = 188
P (card number is not less than 8)
Question 36Question: An organisation selected 2400 families at random and surveyed than to determine a relationship between income level and the number of vehicles in a family. The information gathered is listed in the table below:
Suppose a family is chosen, find the probability that the family chosen is i) earning Rs. 10000 - Rs. 13000 per month and owning exactly 2 vehicles ii) earning Rs. 16000 or more per month and owning exactly 1 vehicle iii) earning less than Rs. 7000 per month and not owning any vehicle iv) earning Rs. 13000 - 16000 per month and owning more than 2 vehicles v) owning not more than 1 vehicle.Answer: Total number of families = 2400
i) Number of families earning Rs. 10000 - Rs. 13000,and owning 2 vehicles = 29
P (the family is earning Rs. 10000 - Rs. 13000 and owns 2 vehicles)
ii) Number of families earning Rs. 16000 or moreand owning 1 vehicle = 579
P (the family is earning Rs. 16000 or more and own 1 vehicle)
iii) Number of families earning less than Rs.7000and not owning any vehicle = 10
P (the family is earning less than Rs. 7000 and does not own any vehicle)
iv) Number of families Rs. 13000 - Rs. 16000and owning more than 2 vehicles =25
P (the family is earning Rs. 13000 - Rs. 16000 and owns more than 2 vehicles)
v) Number of families owning not more than 1 vehicle= 10 + 10 + 1 + 2 + 1 + 160 +305 +535 +469 +579 = 2062
P (the family owns not more than 1 vehicle)
Question 37Question: If E1, E2, E3 cover all possible outcomes of a trial, and P(E1) = 0.7, P (E2) = 0.05. What is the probability of E3?
Statistics and Probability Solutions: Question 38Question: In a survey conducted among 300 students each of whom could speak either Hindi or English or both, it was found that 200 of them could speak Hindi and 250 of them could speak English. If a student is selected at random, what is the probability that he can speak both the languages?Here is a step by step explanation for Statistics and Probability Solutions
Answer: Total number of students = 300
Number of students who can speak both languages
= 200 + 250 - 300 = 150 P (the student can speak both languages)
Problems for Probability with AnswersProblem 1 The box has the 60 caps. In that caps, there are 15 red caps, 10 blue caps and 35 green caps available. Solve the probability with the below outcomes. i) Select the red caps. ii) Select the blue caps.Answer Total caps n(S) = 60 Red caps n (A) =15 Blue caps n (B) = 10 Green caps n(C) =35 i) Assume the P(A) is the probability for choose Red caps .
P(A)=
=
= ii) Assume the P(B) is the probability for choose blue caps.
P(C) =
=
= .We can solve the problems in these method and get the answers.
Problem 2 The bag has the 40 kerchiefs. In that kerchiefs, there are 10 is the gray colour, 15 is the yellow colour and 15 is the white colour. If balu select the white colour kerchief means solve the probability?Answer First list the given information, Total kerchiefs n(S)=40 Gray colour kerchiefs n(A)= 10 Yellow colour kerchiefs n(B)= 15 White colour kerchiefs n(C )= 15 Assume P(E) is the probability for balu select the white colour kerchief.
So P(E)=n(C )/n(S)
=
= .These solve problems with answers are used to study the probability.
Problem 3 Two coins are spinning. What is the probability for if two head are getting?Answer Two coins flipped means the feasible outcomes are {(H,T),(H,H),(T,H),(T,T)} So the total numbers of outcomes are 4. We get the two head outcome is 1.
Therefore the probability is .In the above way we can solve the problems and get the answers.
Example 1: When throwing two dice, what is the probability of getting a sum isa) 10 or 5b) 9 and 5Solution:Let sample space = S, S = {(1, 1), (1, 2), (1, 3) … (6, 5), (6, 6)}, n(S) = 36A be the event of getting a sum is 10, A = {(4, 6), (5, 5), (6, 4)}, n(A) = 3B be the event of getting a sum is 5, B = {(1, 4), (2, 3), (3, 2), (4, 1)} = 4C be the event of getting a sum is 9, C = {(3, 6), (4, 5), (5, 4), (6, 3)} = 4
P(A) = =
P(B) = =
P(C) = =
a) P(A or B) = P(10 or 5) = + =
b) P(B or C) = P(5 or 9) = + = Example 2: A jar contains 7 blue candies and 5 black candies. Two candies are drawn at random. Find the probability of getting both will be blue or black candies.Solution:Let S = Sample space, (7 + 5 = 12).A be the event of getting 2 blue candies.B be the event of getting 2 green candies.(A or B) be the event of getting 2 blue or 2 black candies.
n(S) = 12C2 = = = 66
n(A) = 7C2 = = = 21
n(B) = 5C2 = = = 10
P(A) = =
P(B) = =
P(A or B) = P(A) + P(B) = + =
Learn Probability Problems with Answers - Practice ProblemsProblem 1: A jar contains 6 blue candies and 4 black candies. Two candies are drawn at random with replacement. Find the probability of getting blue and black candies.Problem 2: A jar contains 4 blue balls and 5 black balls. Two candies are drawn at random. Find the probability of getting both will be blue balls.
Answers: 1) 2) Let sample space S = {(1, 1), (1, 2), (1, 3) … (6, 5), (6, 6)}, n(S) = 36A be the event of getting a sum less than 5, A = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (3, 1)}, n(A) = 6B be the event of getting a sum greater than 10, B = {(5, 6), (6, 5), (6, 6)}, n(B) = 3
P(A) = =
P(B) = =
P(A or B) = P(A) + P(B) = + =
Ex 2: A jar has 4 black candies and 7 green candies. Two candies are taken at random. What is the probability of getting both will be black or green color candies.Solution:Let S be the sample space, S = 4 + 7 = 11.A be the event of getting 2 black candies.B be the event of getting 2 green candies.(A or B) be the event of getting 2 black or 2 green candies.We have to select 2 candies from 11 candies.
n(S) = 11C2 = = = 55
n(A) = 4C2 = = = 6
n(B) = 7C2 = = = 21
P(A) = =
P(B) = =
P(A or B) = P(A) + P(B) = + =
P(2 black or 2 green) =
Probability Problems with Answers Help - Practice Problems with AnswersSolve these practice probability problems, it will help you to get practice on how to find the probability.Pr 1: Find the the probability of getting a sum less than 3 or greater than 9 while throwing two dice simultaneously.Pr 2: A jar has 5 white candies and 8 yellow candies. Two candies are taken at random. What is the probability of getting both will be white or yellow color candies.
Answers: 1) 2) Example problem of Expremental probability:A bag contains 10 red marbles, 8 blue marbles and 2 yellow marbles. Find the experimental probability of receiving a blue marble.Solution:Step 1: Take a marble from the bag.Step 2: Record the color and return the marble.Step 3: Repeat a few times (maybe 10 times).Step 4: Count the number of times a blue marble was pick (Suppose it is 6).Step 5: The experimental probability of receiving a blue marble from the bag is 6/10 =3/5
Theoretical Probability Problems - Examples with AnswersExample problem of Theoretical probability:A bag contains 20 marbles. 15 of them are red and 5 of them are blue in color. Find the probability of picking a red marble.Solution:Let’s first answer a few questions here:Step 1: If we are going to randomly select a marble from the bag then what results we have:Step 2: We can also select a red marble or a blue one.Step 3: The next question is what the chances of alternative a red marble are:Step 4: There are 15 red marbles and just 5 blue marbles.Step 5: It’s understandable that we have three times as many red marbles as blue marbles.Step 6: So, the chance of selection a red marble is more than that of the blue one.Step 7: Therefore, the probability of selection a red marble is:Step 8: number of red marbles in the bag / Total number of marbles in the bagStep 9: 15/20=3/4 =0.75 =75%
If a coin is tossed. Find the probability of getting a head.Solution for the problem: In the experiment of tossing a coin once, the number of possible outcomes is two — Head (H) and Tail (T). Let E be the event ‘getting a head’.
The number of outcomes favorable to E, (i.e., of getting a head) is 1 Therefore, P (E) = P (head) = Number of outcomes favorable to E / Number of all possible
outcomes =
Study Probability Example Problem 2:A bag contains a Black ball, a Brown ball and a Purple ball, all the balls being of the same size. David takes out a ball from the bag without looking into it. What is the probability that she takes out the
1. Purple ball?2. Black ball?3. Brown ball?
Solution for the problem: David takes out a ball from the bag without looking into it. So, it is equally likely that she takes out any one of them. Let Y be the event ‘the ball taken out is Purple’, B be the event ‘the ball taken out is Brown’, and R be the event ‘the ball taken out is Black’. Now, the number of possible outcomes = 3. The number of outcomes favorable to the event Y = 1
P(Y) = , P(R) = P (B) = In a rose garden, there are 31 red roses, 24 yellow roses and 15 white roses. Find the probability of the following outcomes? i) Selection of red roses ii) Selection of yellow roses iii) Selection of white rosesSolution: Total number of roses n(S) = 31 + 24 + 15 = 70 Number of red roses n(A) = 31 Number of yellow roses n(B) = 24 Number of white roses n(C) = 15 i) Assume the P(A) is the probability for selection of red roses.
P(A) =
= ii) Assume the P(B) is the probability for selection of yellow roses.
P(B) =
=
= iii) Assume the P(C) is the probability for selection of white roses.
P(C) =
=
=
Example 2 – Normal Probability Online StudyIn a shop, there are 41 products are available. In that, 17 are Anklets, 14 are Bangles and 10 are Rings. Find the probability for the below outcomes? i) Probability for choosing Anklets ii) Probability for choosing BanglesSolution: Total number of products n(S) = 41 Number of Anklets n(A) = 17 Number of Bangles n(B) = 14 Number of Rings n(C) = 10 i) Assume the P(A) is the probability for choosing Anklets.
P(A) =
= ii) Let take P(B) is the probability for choosing Bangles.
P(B) =
= That’s all about normal probability problems online study To find the correct experiment in the following? tossing a coin rolling a die. Both the above. To find the outcome in the following choices? rolling a pair of the dice. Landing on red. Both the above. The probability of choosing a vowel from the alphabet is? 22/26 5/26 1/21
The number from 1 to 10 is chosen at random. find the probability of choosing an odd in them?
1/10 4/10 5/10 These are the examples on study normal probability.t see some of the mathematical probability problems.Choose the letter “g“ from the word “Google”.Solution:The given word is “Google”Here, the letter‘t’ is double time repetitive in the given word.Total letters are, 6Possibility outcomes=2Total number of events in this words=6So the probability,=2/6=1/3=0.33Probability problem 2 with solutions:Choose the letter “t“from the word “tutor”Solutions:The given word is “tutor”Here, the letter‘t’ is two times repetitive in the given statement.Total letters in this given word is “tutor”Possibility outcomes=2Total number of events in this words=5So the probability,=2/5=0.4 These are normal probability problems online tutoring.
Normal Probability 2:Let us we will learn the concept of normal probability help with example problem.In a fridge 25 ice creams. In that fridge 10 chocolate ice creams and 5 butter scotch ice creams and 10 strawberry ice creams. Locate the probability,1. Choose the Chocolate ice creams2. Select the butter scotch ice creamsSolution:2. Imagine the P (A) is the probability for choose the chocolate ice creams.
P (A) =
=
= These are normal probability problems online tutoring.Normal probability problem 2:In a bag there are 50 color files are available. In that bags there are 15 blue files, 15 brown color files, 20 red color files.1. Pick and choose the blue files.2. Pick the Brown files.Solution:Total files n(S) = 50Blue files n (A) = 15Brown files n (B) = 15Red color files n(C) = 201. The P (A) is the probability for select the blue files.
P (A) =
=
= 1. The P (B) is the probability for choose the brown files.
P (B) =
= Normal probability distribution online tutor - Formula:
Z = Where,μ = meanσ = standard deviationX = normal random variable
Normal Probability Distribution Online Tutor – Example Problems:Example 1:If X is a normal random variable with mean and standard deviation calculate the probability of P(X<125). When mean μ = 96 and standard deviation = 35Solution:GivenMean μ = 96Standard deviation σ = 35Using the formula
Z = Given value for X = 125
Z =
= = 0.83Z = 0.83Using the Z table, we determine the Z value = 0.83Z = 0.83 = 0.2967If X is greater than μ then we use this formulaX > μ = 0.5 + Z125 > 96 = 0.5 + 0.2967P(X) = 0.5 + 0.2967 = 0.7967P(X) = 0.7967Example 2:If X is a normal random variable with mean and standard deviation calculate the probability of P(X< 65). When mean μ = 39 and standard deviation = 12Solution:GivenMean μ = 39Standard deviation σ = 12Using the formula
Z = Given value for X = 65
Z =
= = 2.17Z = 2.17Using the Z table, we determine the Z value = 2.17Z = 2.17 = 0.485If X is greater than μ then we use this formulaX > μ = 0.5 + Z65 > 39 = 0.5 + 0.485P(X) = 0.5 + 0.485 = 0.985P(X) = 0.985Find out the given probabilities. P(0 ≤ Z ≤ 2.4) and P(−2.4 ≤ Z ≤ 0). Let we consider z be a standard normal variate.Solution:(i) P(0 ≤ Z ≤ 2.4)
Given: P(0 ≤ Z ≤ 2.4)P (0 ≤ Z ≤ 2.4) = area between the value of z = 0 and the value of z = 2.4 = 0.4918(ii) P(−2.4 ≤ Z ≤ 0)Solution:The given value is P(−2.4 ≤ Z ≤ 0)
P(−2.4 ≤ Z ≤ 0) = P(0 ≤ Z ≤ 2.4). = 0.4918Example problem 2 - solve probbaility and normal distribution:Find out the given probability (i) area to the right of z = 0.8 and the (ii) area to the left of z = 1.6. Let we consider z be a standard normal variate.(i) area to the right of z = 0.8Solution:Given: area to the right of z= 0.8
P(Z > 0.8) = area between the value of z = 0 to the value of z = – area between the value of o to the value of z = 0.8 = P(0 < Z < ) – P(0 ≤ Z < 0.8) = 0.5 – 0.2881 = 0.2119(ii) area to the left of z = 1.6Solution:
Given : area to the left of z = 1.6
= P(Z < 1.6) = 0.5 + 0.4452 = 0.9452Example1:The average life of a certain kind of coast is 10 years, with a standard deviation of 2 years. If the producer is willing to replace only 3% of the motors that fail, how long a guarantee should he offer?Solution:X=motor lifex= guarantee period
We need to find the value (in years) that will give us the bottom 3% of the distribution. These are the motors that we are willing to replace under the guarantee.P(X<x) = 0.03From table get 0.5 values0.5 - 0.03 = 0.47 after that value of Z is -1.88 Z = (X – μ) / σ(x-10)/2 = -1.88 the value of x is6.24The guarantee time is 6.24 years.Example 2:A corporation pays its employees an average salary of $3.25 an hour with a standard deviation of 60 cents. If the wages are around normally distributed, determinei) the proportion of the workers getting wages between 2 75and3.69 an hourii) The minimum wage of the highest 5%.
Solution:i) The proportion of the workers getting wages between 2 75and3.69 an hourZ1 = ((2.75-3.25)/0.6) = -0.833 Z2 = ((3.69-3.25)/0.6) = 0.733P (2.75< X<3.69) = P (-0.833< Z< 0.733) = 0.298 + 0.268 = 0.566 Totally 56.6% of the workers have wages.ii) The minimum wage of the highest 5%.X=1.645 from normal distribution table(X-3.25)/0.6 = 1.645After evaluation the value of X = 4.237The minimum salary of the top 5% if salary is $4.24
Example 1: Mean length of 200 parts produced by a company was 10.05 mm with a standard deviation of 0.02 mm. What is the probability that a part chosen at random would have a length(a) between 10.03 mm and 10.08 mm?(b) between 10.06 mm and 10.07 mm?Solution:X = length of part
If we know μ and σ, then
(a) 10.03 is standard deviations below the mean
10.08 is standard deviations above the meanP(10.03 < X < 10.08= (-1 < Z < 1.5)= 0.3413 + 0.4332 = 0.7745So the probability is 0.7745.
(b) 10.06 is standard deviations above the mean;
10.07 is standard deviation above the meanP(10.06 < X < 10.07)= P(0.5 < Z < 1)= 0.3413 – 0.1915= 0.1498So the probability is 0.1498.
Probability of Normal Distribution Learning - Learning Example 2Example 2: Mean height of 50 parts produced by a company was 10.05 cm with a standard deviation of 0.02 cm. What is the probability that a part chosen at random would have a heighta) less than 10.01 cm?
b) greater than 10.09 cm?Solution:
(a) 10.01 is standard deviation below the mean.P(X < 10.07)= P(Z < -2)= 0.5 – 0.4792= 0.0228.So the probability is 0.0228.(b) 10.09 is 2 standard deviation above the meanP(X > 10.09) = 0.0228The formula for the Poisson distribution is,
f(x) =
where, x is the Poisson random valueis the average rate of success
Now, we are going to see some of the problems on poisson distribution questionstutor help. From this help, we can get clear idea about poisson distribution questions.
Example Problem-poisson Sistribution Questions Tutor:Example problem 1:Today, 4 foreigners visited the city Washington. Find the possibilities for exactly 7 foreigners to be visit on tomorrow. Use Poisson distribution to solve.Solution:Step1:Find the value of e-λ.Where, λ=4 and e=2.718
e-λ = (2.718)-4 = =0.01832Step2: Find the value of λx.
Where, λ=4 and x=7λx = 47 = 16384Step3:Find f(x).
= = 0.05955So, the probability of exactly 7 foreigners to visit on tomorrow is 0.05955.
Additional Problem-poisson Sistribution Questions Tutor:Example problem 2:
If λ= 2 and x = 6, then find the poisson distribution.Solution:Step1:Find the value of e-λ.Where, λ=2 and e=2.718
e-λ = (2.718)-2 = =0.135363Step2: Find the value of λx.
Where, λ=2 and x=6λx = 26 = 64Step3:Find f(x).
= = 0.012So, the poisson distribution is 0.012
Practice Problems-poisson Sistribution Questions Tutor:Problem 1:If λ= 2 and x = 5, then find the poisson distribution.(solution: 0.036) Problem 2:If λ= 5 and x = 10, then find the poisson distribution.(Solution: 0.018).
The formula for the Poisson distribution is,
f(x) =
where, x is the Poisson random valueis the average rate of success
Now, we are going to see some of the problems on poisson distribution questionstutor help. From this help, we can get clear idea about poisson distribution questions.
Example Problem-poisson Sistribution Questions Tutor:Example problem 1:Today, 4 foreigners visited the city Washington. Find the possibilities for exactly 7 foreigners to be visit on tomorrow. Use Poisson distribution to solve.Solution:Step1:Find the value of e-λ.Where, λ=4 and e=2.718
e-λ = (2.718)-4 = =0.01832Step2: Find the value of λx.
Where, λ=4 and x=7λx = 47 = 16384Step3:Find f(x).
= = 0.05955So, the probability of exactly 7 foreigners to visit on tomorrow is 0.05955.
Additional Problem-poisson Sistribution Questions Tutor:Example problem 2:If λ= 2 and x = 6, then find the poisson distribution.Solution:Step1:Find the value of e-λ.Where, λ=2 and e=2.718
e-λ = (2.718)-2 = =0.135363Step2: Find the value of λx.
Where, λ=2 and x=6λx = 26 = 64Step3:Find f(x).
= = 0.012So, the poisson distribution is 0.012
Practice Problems-poisson Sistribution Questions Tutor:Problem 1:If λ= 2 and x = 5, then find the poisson distribution.(solution: 0.036) Problem 2:If λ= 5 and x = 10, then find the poisson distribution.
(Solution: 0.018).
Example 1: Solve Poisson distribution where, = 2, x = 5 and e = 2.718Tutoring Solution: Step 1: Given:
= = 0.0025Tutoring Result: Poisson Distribution = 0.0025
Practice Problems to Poissons Distribution Online TutoringExample 1: Solve Poisson distribution where, = 12, x = 14 and e = 2.718.Tutoring Answer: 0.09Example 2: Solve Poisson distribution where, = 15, x = 17 and e = 2.718.Tutoring Answer: 0.085A manufacturer of cotton pins knows that 4% of his product is defective. If he sells pins in boxes of 100 and guarantees that not more than 2 pins will be defective. Find the approximate probability that a box will fail to meet the guaranteed quality.Solution:
The value of p is p = , n = 100
The mean value is lambda = n p = ( ) (100) = 4By the Poisson distribution
P[X = x] = Probability that a box will to meet the guaranteed quality = P[X > 2] P[X > 2] = 1- P[X ≤ 2] P[X > 2] = 1- (P (0) +P (1) +P (2))
P[X > 2] = 1- (1 + 4+ ( )) P[X > 2] = 1- (1+4+8) P[X > 2] = 1- (13) P[X > 2] = 1- 0.0183(13) P[X > 2] = 1- 0.2379 P[X > 2] = 0.7621The probability for the box will fail to meet the guaranteed quality is 0.7621.Example 2 for Poisson distribution football online tutor:A car- hire firm has three cars. The number of demands for a car on each day is distributed as a Poisson distribution with mean of 1.8. Calculate the proportion of days on which neither car is used and the proportion of days on which some demand is refused.Solution:Let X is the number of demands for a car.The given mean value is 1.8.By the Poisson distribution
P[X = x] = Proportion of the days on which neither car is used = P[X = 0] = = 0.1653
Proportion of days on which some demand is refused = P[X > 3] P[X > 3] = 1- P[X ≤ 3] P[X > 3] = 1- [P (0) +P (1) + P (2) +P (3)] P[X > 3] = 1- (1+ 1.8+ 1.62+ 0.972) P[X > 3] = 1 - (0.1653) (5.392) P[X > 3] = 1 - 0.8913 P[X > 3] = 0.1087The proportion of days on which neither the car used is 0.1653. The proportion of days on which some demand refused is 0.1087.
Explanation to Mean of Poisson Distribution Tutoring:The explanation for the Poisson distribution mean are given below the following section,
Formula:
Mean of Poisson distribution = E(Y) = x P(Y)where, E(X) = Mean of Poisson distribution P(Y) = Probability of Poisson distribution
= Sum of all values of Poisson
Example Problems to Mean of Poisson Distribution Tutoring:Problem 1- Mean of Poisson distribution tutoring
Calculate the mean for the following Poisson distribution table,x 3 5 7 8 11
P(Y)
Solution:Step 1: Given:x = 3, 5, 7, 8, 11
P(Y) = , , , , Step 2: Formula:
Mean of Poisson distribution = E(Y) = x P(Y)Step 3: Find:
E(X) = 3( ) + 5( ) + 7( ) + 8( ) + 11( )
= + + + +
= 0.75 + 2.5 + 1.4+ 2.66 + 5.5 = 12.81Result: Mean of Poisson distribution = 12.81Problem 2 - Mean of Poisson distribution tutoringCalculate the mean for the following Poisson distribution table,
x 1 4 5 6 7
P(Y)
Solution:Step 1: Given:x = 1, 4, 5, 6, 7
P(Y) = , , , , Step 2: Formula:
Mean of Poisson distribution = E(Y) = x P(Y)Step 3: Find:
E(X) = 1 ( ) + 4( ) + 5( ) + 6( ) + 7( )
= 1/2+ 1+ + + = 0.5 + 1 + 1.66 + 1.2 + 1.16 = 5.96Result: Mean of Poisson distribution = 5.96
Practice Problems to Mean of Poisson Distribution Tutoring:Problem 1 - Mean of Poisson distribution tutoring
Calculate the mean for the following Poisson distribution table,x 1 3 4 5 9
P(Y)
Answer: 8.08Problem 2 - Mean of Poisson distribution tutoringCalculate the mean for the following Poisson distribution table,
x 2 5 6 8 10
P(Y)
Answer: 8.27
Binomial distribution:It describes the number of successes occurring from 'n' number of experiments.Formula for binomial distribution:P(x = r) = nCr pr (1 – p)n-r
Where,n – number of eventsr – Number of successp – Probability of success
nCr - combination = Poisson distribution:It describes the number of events happened in a fixed time.
Formula for Poisson distribution:
f(x) = where,
– Average ratex – Poisson random variablee – base of logarithm = 2.718
Example Problems to Binomial and Poisson Distribution Tutoring:Example: 1 A shop has 150 computers; each computer has 4% probability of not working. If Ram wants to buy 6 computers then calculate the likelihood that 3 will be damaged.Solution:Given:n = 6r = 3
p = 4% = = 0.04Formula for binomial distribution:P(x = r) = nCr pr (1 – p)n-r
P(x = 3) = 6C3 0.043 (1 – 0.04)6-3
Step 1:
6C3 =
= = = 20Step 2:0.043 = 0.000064Step 3:
(1 – 0.04)6-3 = (0.96)3
= 0.884736Step 4:P(x = 3) = (20)(0.000064)(0.884736)= 0.0153Answer: 0.0011Example: 2In a coaching center, 5 college students arrived today. Determine the possibility for 7 college students will be arrived on the next day.Solution:Given:
= 5x = 7Step 1:
Poisson distribution = Step 2:e-5 = 0.006738Step 3:
= (5)7 = 78125Step 4:
=
= = 0.1044Answer: 0.1044
Practice Problems to Binomial and Poisson Distribution Tutoring:Problem: 1 A mobile shop has 500 cell phones and each mobile has 5% probability of not working. Ravi want to buy 4 mobiles then determine the likelihood that 1 will be damaged.Answer: 0.171Problem: 2 In a class, 6 school students are absent Friday. Determine the possibility for 12 school students will be absent on Saturday.Answer: 0.011Example: In a college there are 5 staffs absent today. Find the possibilities for exactly 7 staff to be taking leave on tomorrow.Solution:Step1:Find e-λ.
== 0.1044921875.Hence there are 10% possibilities for 5 staffs to be taking leave of college on tomorrow. Example: In a college there are 5 staffs absent today. Find the possibilities for exactly 6 staff to be taking leave on tomorrow.Solution:Step1:Find e-λ.Where, λ=5 and e=2.718
= = 0.1462890.Hence there are 0.14% possibilities for 6 staffs to be taking leave of an office on tomorrow. Example:
In an office 4 clients visit today. Find the possibilities for exactly 7 clients to be visit on tomorrow.Solution:Step1:Find e-λ.Where, λ=4 and e=2.718
= = 0.05956.Hence there are 6% possibilities for 7 clients to be visit an office on tomorrow.
Poisson Binomial Distribution Online Tutoring Practice Problem:Problem: In an office there are 4 staffs absent today. Find the possibilities for exactly 6 staff to be taking leave on tomorrow.Answer:0.1042388.There are 10% possibilities for 6 staffs to be taking leave of an office on tomorrow.Problem: In an office there are 5 staffs absent today. Find the possibilities for exactly 7 staff to be taking leave on tomorrow.Answer:0.104499.There are 0.10% possibilities for 7 staffs to be taking leave of an office on tomorrow.Example: 1 A shop has 100 televisions; each television has 5% probability of not working. If you want to select 10 televisions randomly then find the likelihood that 2 will be broken.Solution:Given:n = 10r = 2
p = 2% = = 0.02Formula for binomial distribution:P(x = r) = nCr pr (1 – p)n-r
= 0.850Step 4:P(x = 2) = (45)(0.0004)(0.850)= 0.0153Answer: 0.0153Example: 2In a library, 3 students arrived today. What is the possibility for 5 students will be arrived on next day.Solution:Given:
= 3x = 5Step 1:
Poisson distribution = Step 2:e-3 = 0.04978Step 3:
= (3)5 = 243Step 4:
=
= =
Answer: 0.101Example Problems to Binomial and Poisson Distribution Help:Problem: 1 A cell shop has 200 cell phones; each cell phone has 3% probability of not working. If you want to select 5 cell phones randomly then find the likelihood that 1 will be broken.Answer: 0.133Problem: 2In coaching center, 5 students are absent today. What is the possibility for 10 students will be absent on next day.Answer: 0.018
Solve Poisson Binomial Distribution Example Problem:Example: In an office there are 4 staffs absent today. Find the possibilities for exactly 6 staff to be taking leave on tomorrow.Solution:Step1:Find e-λ.Where, λ=4 and e=2.718e-λ = (2.718)-4 = 1/54.575510850576= 0.0183232366Step2: Find λx.
Where, λ=4 and x=6.λx = 46 = 4096.Step3:Find f(x).f(x) = e-λλx / x!f(6) = (0.0183232366)(4096) / 6! .=75.05195 /720= 0.1042388.Hence there are 10% possibilities for 6 staffs to be taking leave of an office on tomorrow. Example: In an office there are 5 staffs absent today. Find the possibilities for exactly 7 staff to be taking leave on tomorrow.Solution:Step1:Find e-λ.Where, λ=5 and e=2.718e-λ = (2.718)-5 = 1/148.3362= 0.006741Step2: Find λx.
f(7) = (0. 006741)(78125) / 7! .= 526.67521 /5040= 0.104499.Hence there are 0.10% possibilities for 7 staffs to be taking leave of an office on tomorrow.Example:
In an office 2 clients visit today. Find the possibilities for exactly 4 clients to be visit on tomorrow.Solution:Step1:Find e-λ.Where, λ=2 and e=2.718e-λ = (2.718)-2 = 1/7.387524= 0.135633Step2: Find λx.
Where, λ=2 and x=4.λx = 24 = 16.Step3:Find f(x).f(x) = e-λλx / x!f(4) = (0. 135633)(16) / 4! .= 2.16581/24= 0.09024.Hence there are 9% possibilities for 4 clients to be visit an office on tomorrow.
Solve Poisson Binomial Distribution Practice ProblemProblem:In an office 3 clients visit today. Find the possibilities for exactly 5 clients to be visit on tomorrow.Answer:0.100084.There are 10% possibilities for 5 clients to be visit an office on tomorrow.Problem:In an office 4 clients visit today. Find the possibilities for exactly 5 clients to be visit on tomorrow.Answer:0.156358.There are 15% possibilities for 5 clients to be visit an office on tomorrow.
