Note 2 probability

PROBABILITY THEORY CN303/1/ 1

PROBABILITY THEORY OBJECTIVES General Objective

� To understand the concept of probability. Specific Objectives At the end of the unit you should be able to:

� Define Classical Probability and state the concept of Experiments and Events

� Define the following events:

• Conditional Events • Independent Events • Mutually Exclusive Events

� List the elements in the sample spaces To find the probability of an event based on Classical probability Use the set theory to explain:

• Venn diagrams • complementary of sets • Union of sets • Intersection of sets • Null sets

UNIT 1


Probability as a general concept can be defined as the chance of an event occurring. Probability is the basis of inferential statistics. For example, predictions are based on probability, and hypotheses are tested by using probability.

1.0 INTRODUCTION

So much in people’s lives is affected by chance. From the time a person awakes until he or she goes to bed, that person makes decision regarding the possible events that are governed at least in part by chance. For example, should I carry an umbrella to class today?

Will my motorcycle battery last until the end of the semester? Should I accept a new job?

INPUT


1.1 PROBABILITY � A probability experiment is a chance process that leads to well-defined

results called outcomes. � An outcome is the result of a single trial of a probability experiment. � A sample space is the set of all possible outcomes of a probability

experiment. � An event consists of the outcomes of a probability experiment.

Some sample spaces for various probability experiments are shown here.

Experiment Sample space Toss one coin Head, Tail Roll a die 1, 2, 3, 4, 5, 6 Answer a true-false question True, False Toss two coins Head-head, tail-tail, head-tail, tail-head

Example 1.1 1. a) Find the sample space for rolling two dice.

b) Find the sample space for the gender of the children if a family has three children. Use B for boy and G for girl.

2. Use a tree diagram to find the sample space for the gender of three children in a family, as in ACTIVITY 1B


Solution to Example 1.1 1. a

Since each die can land in six different ways, and two dice are rolled, the sample space can be presented by a rectangular array. The sample space is the list of pairs of numbers in the chart.

Die 1 Die 2

1 2 3 4 5 6 1 (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) 2 (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) 3 (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) 4 (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) 5 (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) 6 (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6)

1. b

There are two genders, male and female, and each child could be either gender. Hence there are eight possibilities, as shown here.

BBB BBG BGB GBB GGG GGB GBG BG G

2.

children

B G

B G B G

B G B G B G B G


ACTIVITY 1A

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INTPUT…!

1. Find the sample space for tossing two coins. 2. A die is rolled and a coin is tossed. Show the sample space.


FEEDBACK TO ACTIVITY 1A

1. S = {(H, H), (H, T), (T, H),(T,T)} 2.

Die 1 2 3 4 5 6 Coin Head (H))

(H, 1) (H, 2) (H, 3) (H, 4) (H, 5) (H, 6)

Tail (T) (T, 1) (T, 2) (T, 3) (T, 4) (T, 5) (T, 6)


1.2 CERTAIN AND COMPLEMENTARY EVENTS When a die is rolled, the sample space is S = {1, 2, 3, 4, 5, 6}. Now let us define event A as ‘number 1 appears on the die’s surface, therefore complement of A (written as A’) consists of all the number on the die’s surface excluding 1, therefore A’ = {2, 3, 4, 5, 6}. Events can be presented pictorially by Venn diagrams. Figure (a) shows a simple event E. The area inside the rectangle represents all the events in the sample space(S). Figure (b)

Shows the complement of an event (−E ), which is the area inside the rectangle

but outside the circle representing E.

Fig a Fig b

S Eee

E

- S E

E

INPUT


1.2.1 SET DESIGNATIONS

1. Sample space, S is represented by elements in a rectangle. Any event, E

is represented by its elements in a circle.

2. E’ or −E is the complement of E. E’ means event E never occurred.

3. 21 EE ∪ means either E1 or E2 or both have occurred.

S

E

E’

E

S

E1 E2


4. 21 EE ∩ means both occurred.

5. E1 and E2 are two mutually exclusive events in which 21 EE ∩ =φ . They have no shared outcomes.

6. nEEEE .....,, ,321 are mutually exclusive and finite if and only if

i) φ=∩ ji EE for every i and j,

ii) SEEEE n =∪∪∪∪ ......321

E1 E2 E4 E3 E5

21 EE ∩

S

1E 2E S


1.2.2 SET IDENTITIES The following identities can be used if there is a need. 1. AAA =∪ 8. ABBA ∩=∩ 2. φφ =∪A 9. )()()( CABACBA ∪∩∪=∩∪ 3. SSA =∪ 10. )()()( CABACBA ∩∪∩=∪∩ 4. AAA =∩ 11. '')'( BABA ∩=∪ 5. ABBA ∪=∪ 12. '')'( BABA ∪=∩ 6. ASA =∩ 13. )'( BAABA ∩∪=∪ 7. φφ =∩A 14. )'()( BABAB ∩∪∩= Example 1.2

1. A = {1, 2, 3, 4, 5, 6} and B = {1, 3, 6, 7}, find: i) BA ∩ ii) BA ∪ iii) n( )BA ∪ 2. A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 6} and C = {3, 5, 7, 9}, find: i) BA ∩ ii) CA ∩ iii) CB ∩ iv) CBA ∩∩ v) n )( CBA ∪∪


Solution to example 1.2

1. i) BA ∩ = {1, 3, 6}

ii) BA ∪ = {1, 2, 3, 4, 5, 6, 7} iii) n )( BA ∪ = n(A) + n(B) – n )( BA ∩ = 6 + 4 – 3 = 7 *If A, B and C are finite sets, therefore:

)()()()()()()()( CBAnCBnCAnBAnCnBnAnCBAn ∩∩+∩−∩−∩−++=∪∪

2. i) BA ∩ = {2, 4, 6}

ii) CA ∩ = {3, 5} iii) CB ∩ = { }φ iv) CBA ∩∩ = { }φ v) n )( CBA ∪∪ )()()()()()()( CBAnCBnCAnBAnCnBnAn ∩∩+∩−∩−∩−++

= 6 + 3 + 4 – 3 – 2 – 0 + 0 = 8


ACTIVITY 1B TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

1. Find the sample space for tossing two coins. 2. Nasir normally has one type of drink after lunch everyday. He randomly

drinks tea, coffee or simply plain water. If event A represents ‘Mamat has one type of drink after lunch’, list down the elements in the sample space S and event A. Find the relationship between the sample space and the A set.

3. If n (A ∪ B) = 45, 5)( =∩ BAn , and n(B) = 22, find n(A).


FEEDBACK TO ACTIVITY 1B 1. S = {HH, HT,TH,TT) 2. S={tea, coffee, water}, A = {tea, coffee, water}, therefore S = A 3. 28


1.3 CLASSICAL PROBABILITY Classical probability uses sample spaces to determine the numerical probability that an event will happen. One does not actually have to perform the experiment to determine that probability. Classical probability assumes that all outcomes in the sample space are equally likely to occur. For example, when a single die is rolled, each outcome has the same probability of occurring. Since there are six outcomes, each outcome has a probability of 6

1 .

Equally likely events are events that have the same probability of occurring

Formula for Classical Probability The probability of any event E is

_____Number of outcomes in E____________ Total number of outcomes in the sample space

This probability is denoted by

)(

)()(

Sn

EnEP =

This probability is called classical probability , and it uses the sample space S.

INPUT


Probability Rule 1 The probability of any event E is a number (either a fraction or decimal) between and including 0 and 1. This is denoted by 1)(0 ≤≤ EP . Probability Rule 2 If an event E cannot occur (i.e., the even contains no members in the sample space), the probability is zero. Probability Rule 3 If an event is certain, the probability of E = 1. Probability Rule 4 The sum of the probabilities of the outcomes in the sample space is 1

Example 1.3 1. If a family has three children, find the probability that all the children are girls. 2. When a single die is rolled, find the probability of getting a 9. 3. When a single die is rolled, what is the probability of getting a number less than 7?

The next example illustrates the probability rules


Solution to Example 1.3 1. The sample space for the gender of children for a family that has three

children is BBB, BBG, BGB, GBB, GGG, GGB, GBG, and BGG. (see t6he tree diagram in the basic concept’s section). Since there is one way in eight possibilities for all three children to be girls,

P(GGG) = 8

1 2. Since the sample space is 1, 2, 3, 4, 5, and 6, it is impossible to get a 9.

Hence, the probability is P(9) = 060 =

3. Since all outcomes, 1, 2, 3, 4, 5, and 6, are less than 7, the probability is P(number less than 7) = 16

6 = ** (Rule 4) For example, in a roll of a fair die, each outcome in the sample space has a probability of 6

1 . Hence, the sum of the probabilities of the outcomes is as shown.

Outcome 1 2 3 4 5 6 Probability

61 6

1 61 6

1 61 6

1

Sum 61 + 6

1 + 61 + 6

1 + 61 + 6

1 =1


ACTIVITY 1C TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 1. What is the probability of throwing a number greater than 4 with a die whose

faces are numbered from 1 to 6? 2. In a competition a prize is given for correctly forecasting the results of six football matches. If a competitor sends in ten different forecasts, what is the probability, that he receives the prize? 3. a) One red and one black marble are concealed in a bag. Find the probability of drawing a red marble.

b) When three red and one black marbles are placed in the bag, find the probability of drawing one red marble. What is the probability of drawing one black marble?

4. A box contains 132 rivets of which 32 are undersized, 47 are oversized and

62 are satisfactory. Determine the probability of drawing at random: (a) one undersized; (b) one oversized; and

(c) one satisfactory rivet from the box.

5. Four hundred resistors are examined and 6% are found to be defective.

Determine the probability that one selected at random will be defective and also the probability that it will not be defective.

6. A purse contains 7 copper and 13 silver coins. Determine the probability of

selecting a copper coin when one is taken at random. 7. Determine the probability of winning a prize in a raffle by buying 3 tickets,

when there are 7 prizes and a total of 450 tickets sold. 8. Determine the probability of an event not happening when the probability of it

happening is 7/93.


FEEDBACK TO ACTIVITY 1C

1. 1/3 or 0.33 or %3331

2. 0.0137 or 1.37%

3. (a) ½, (b) ¾ and ¼

4. (a) 132

23 (b)

132

47 (c)

66

31

5.

50

47,

30

3

6. 20

7

7. 7/150

8. 86/93


1.4 COMPLEMENTARY EVENTS

Formula for empirical probability Given a frequency distribution, the probability of an event being in a given class is

P(E)=n

f= , of which f is frequency for the class and n is the total

frequencies in the distribution

The complement of an event E is the set of outcomes in the sample space that are not included in the outcomes of event E. The complement of E is

denoted by E’ or −E (read “E bar”).

Rule for complementary events

)(1)( EPEP −=−

INPUT


Example 1.4

1. 25 students were asked if they like this module. The responses were classified as “yes”, “no”, or “undecided”. The results were categorized in a frequency distribution, as shown. Find the probability that a person responded “no”.

Response Frequency Yes 15 No 8 Undecided 2 Total 25

2. In a sample of 50 students, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had type AB blood. Set up a frequency distribution and find the following probabilities:

a. A student has type O blood b. A person has type A or type B blood c. A person has neither type A nor type O blood d. A person does not have type AB blood

3. Hostel records indicated that students stayed in the hostel for the number

of days during school break shown in the distribution.

Number of days stayed

Frequency

3 15 4 32 5 56 6 19 7 5

Total 127

Find the probabilities.

a. A student stayed exactly 5 days. b. A student stayed less than 6 days c. A student stayed at most 4 days. d. A student stayed at least 5 days.


4. Based on the Venn diagram below, A and B are two events in the sample space S. Find:

a. P(A) (g) )'( BAP ∩ b. P(A’) (h) )''( BAP ∩ c. P(B) (i) )( BAP ∪ d. P(B’) (j) )'( BAP ∪ e. )( BAP ∩ (k) )'( BAP ∪ f. )'( BAP ∩ (l) )''( BAP ∪

S 40 aAr

B A 35 5 20


Solution to Example 1.4

1. P(E)= 258=n

f 2.

i. P(O) = 50

21=n

f

ii. P(A or B) = 5027

505

5022 =+ (Add the frequencies of the two

classes) iii. P(neither A nor O) = 50

7502

505 =+ (Neither A nor O means tat a

student has either type B or type AB blood). iv. P(not AB) = 1 – P(AB) = 25

245048

5021 ==−

3. a) P(5) = 12756

b) P(less than 6 days) = 127103

12756

12732

12715 =++ (Less than

6 days means either 3, or 4 or 5 days.) c) P(at most 4 days) = 127

4712732

12715 =+ (At most 4 days means 3 or 4

days.

d) P(at least 5 days) = 12780

1275

12719

12756 =++ (At least 5 days means

either 5, or 6, or 7 days.)

Type Frequency A 22 B 5

AB 2 O 21

Total 50


4 . n(S) = 35 + 5 + 20 + 40

a. n(A) = 40 and P(A) =5

2

100

40

)(

)( ==Sn

An

b. n(A’) = 60 and P(A’) = 5

3

100

60

)(

)'( ==Sn

An

c. n(B) = 25 and P(B) = 4

1

100

25

)(

)( ==Sn

Bn

d. n(B’) = 75 and P(B’) = 4

3

100

75

)(

)'( ==Sn

Bn

e. n(A ∩ B) = 5 and P(A ∩ B) =20

1

100

5

)(

)( ==∩Sn

BAn

f. n(A ∩ B’) = 35 and P(A B∩ ’) = 20

7

100

35

)(

)'( ==∩Sn

BAn

g. n(A’ )B∩ = 20 and P(A’∩ B) = 5

1

100

20

)(

)'( ==∩Sn

BAP

h. n(A’ ∩ B’) = 40 and P(A’ ∩ B’) = 5

2

100

40

)(

)''( ==∩Sn

BAP

i. n(A ∪ B) = 60 and P(A ∪ B) = 5

3

100

60

)(

)( ==∪Sn

BAP

j. n(A ∪ B’) = 80 and P(A ∪ B’) = 5

4

100

80

)(

)'( ==∪Sn

BAP

k. n(A’ )B∪ = 65 and P(A’ )B∪ = 20

13

100

65

)(

)'( ==∪Sn

BAP

l. n(A’ )'B∪ = 95 and P(A’∪ B’) = 20

19

100

95

)(

)''( ==∪Sn

BAn


ACTIVITY 1D

TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

1. If there are 50 tickets sold at a raffle and on person buys 7 tickets, what is the probability of that person winning a price?

2. A survey found that 53% of Polytechnic students think this module is the best

of all the modules ever published. If a student is selected at random, find the probability that he or she will disagree or have no opinion at all.

3. A couple has three children. Find each probability.

a) Of all boys b) Of all girls c) Of exactly two boys or two girls d) Of at least one child of each gender


FEEDBACK TO ACTIVITY 1D 1. 50

7 2. 47% 3 a. 8

1 b. 41 c. 4

3 d. 43


1.5 THE ADDITION RULES FOR PROBABILITY Many problems involve finding the probability of two or more events. For example, in your class gathering, one might wish to know, for a person selected at random, that a person is a female or is wearing glasses. In this case there are three possibilities to consider.

1. The person is a female. 2. The person is wearing glasses. 3. The person is a female and she is wearing glasses.

Consider another example. In the same gathering there are male and female students. If a person is selected at random, what is the probability that the person is a male or a female student? In this case, there are only two possibilities:

1. The person is a female. 2. The person is a male.

The difference between the two examples is that in the first case, the person selected can be a female and is wearing glasses at the same time. In the second case, the person selected cannot be both a female and a male at the same time. In the second case, the two events are said to be mutually exclusive; in the first case, they are not mutually exclusive.

INPUT


Two events are mutually exclusive if they cannot occur at the same time (i.e., they have no outcomes in common) Addition Rule 1 When two events A and B are mutually exclusi ve, the probability that A and B will occur is

P(A or B) =P(A) + P(B)

Addition Rule 2 If A and B are not mutually exclusive , then P(A or B) = P(A) + P(B) – P(A ∩ B) Example 1.5 1. A restaurant has 3 pieces of chicken karipap, 5 pieces of potato karipap and 4 pieces of fish karipap. If a customer selects a piece of karipap for dessert, find the probability that it will be either potato or fish. 2. There are 20 buffaloes, 13 cows and 6 goats in a lorry. If an animal is selected at random, find the probability that that animal is either a cow or a goat. 3. A day of the week is selected at random; find the probability that it is a weekend day (Saturday or Sunday). 4. In a hospital unit there are eight nurses and five doctors. Seven nurses and three doctors are females. If a staff person is selected the probability that the subject is a nurse or a female.


Solution to Example 1.5 1. Since there are 12 pieces of karipap, P(potato or fish) = P(potato) + P(fish) = 4

3129

124

125 ==+

The events are mutually exclusive. 2. P(cow or goat) = P(cow) + P(goat) = 39

19396

3913 =+

3. P(Saturday or Sunday) = P(Saturday) + P(Sunday) = 7

271

71 =+

4. The sample space is shown below:

Staff Females Males Total Nurses 7 1 8 Doctors 3 2 5 Total 10 3 13

The probability is P (nurse or male) = P(nurse) + P(male) – P(male nurse) = 13

10131

133

138 =−+


ACTIVITY 1E TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…! 1. At a convention there are seven mathematics instructors, five computer science instructors, three statistics instructors, and four science instructors. If an instructor is selected, find the probability of getting a science instructor or a math instructor. 2. In a statistics class there are 18 juniors and 10 seniors; 6 of the seniors are females, and 12 of the juniors are males. If a student is selected at random, find the probabilities of selecting the following: a. a junior or a female b. A senior or a female c. A junior or a senior 3. A woman’s clothing store owner buys from three companies: A, B, and C. The most recent purchases are shown here.

Product A B C Dresses 24 18 12 Blouses 13 36 16

If one item is selected at random, find the following probabilities. a. It was purchased from company A or is a dress. b. It was purchased from company B or company C. c. It is a blouse or it was purchased from company A.


4. A grocery store employs cashiers, stock clerks, and deli personnel. The distribution of employees according to marital status is shown next.

Marital status

cashiers Stock clerks

Deli personnel

Married 8 12 3

Not married

5 15 2

If an employee is selected at random, find these probabilities: a. The employee is a stock clerk or married. b. The employee is not married. c. The employee is a cashier or is not married. 5. RTM, TV3, and NTV7 have quiz shows, comedies, and dramas. The number of each is shown here.

Type of show

RTM Tv3 Ntv7

Quiz show

5 2 1

Comedy 3 2 8 Drama 4 4 2

If a show is selected at random, find these probabilities. a. The show is a quiz show or it is shown on TV3. b. The show is a drama or a comedy. c. The show is shown on NTV7 or it is a drama.


FEEDBACK TO ACTIVITY 1E

1. 19

11

2. a. 7

6 b.

7

4 c. 1

3. a. 118

67 b.

118

81 c.

59

44

4. a. 45

38 b.

45

22 c.

3

2

5. a. 31

14 b.

31

23 c.

31

19


1.6 THE MULTIPLICATION RULES AND CONDITIONAL PROBABILITY The previous section showed that the addition rules are used to compute probabilities for mutually exclusive and not mutually exclusive events. This section introduces two more rules, the multiplication rules. These rules can be used to find the probability of two or more events that occur in sequence. For example, if a coin is tossed and then a die is rolled, one can find the probability of getting a head on the coin and a 4 on the die. These two events are said to be independent since the outcome of the first event (tossing a coin) does not affect the probability outcome of the second event (rolling a die).

In order to find the probability of two independent events that occur in sequence, one must find the probability of each event occurring separately and then multiply the answers. For example, if a coin is tossed twice, the probability of getting two heads is 4

121

21 . = . The result can be verified by looking at the sample space, HH,

HT, TH, and TT. Then P (HH) = 41 .

Two events A and B are independent if the fact that A occurs does not affect the probability of B occurring.

INPUT


Multiplication Rule 1 When two events are independent, the probability of both occurring is P(A and B) = P(A).P(B)

Example 1.6 1. A coin is flipped and a die is rolled. Find the probability of getting a head on the coin and a 4 on the die. 2. An urn contains three red balls, two blue balls, and five white balls. A ball is selected and its color noted. Then it is replaced. A second ball is selected and its color noted. Find the probability of each of the following. a. selecting two blue balls. b. Selecting a blue ball and then a red ball. c. Selecting a red ball and then a blue ball. 3. A pool found that 46% of students say they have suffered great stress at least once in the exam week. If three students are selected at random, find the probability that all three will say that they suffer great stress al least once in the exam week. Solution to Example 1.6 1. P(head and 4) = P(head).P(4) = 12

161

21 . = , note that the sample space for the

coin is H, T; and for the die is 1, 2, 3, 4, 5, 6. 2. a. P(blue and blue) = P(blue).P(blue) = 25

1100

4102

102 . ==

b. P(blue and white) = P(blue).P(white) = 101

10010

105

102 . ==

c. P (red and blue) = P(red).P(blue) = 503

1006

102

103 . ==

3. Let S denote stress, then P(S and S and S) = P(S).P(S).P(S) = (0.46)(0.46)(0.46) =0.097


ACTIVITY 1F TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITH THE NEXT INPUT…!

1. Two balls are drawn in turn with replacement from a bag containing 8 red balls, 15 white balls, 24 black balls and 17 orange balls. Determine the probabilities of having:

(a) two red balls; (b) a red and a white ball; (c) no orange balls; (d) a black and red or black and orange ball; (e) at least one black ball; (f) a white ball on the first draw but the second ball not white.

2. The probability of three events happening are 1/8 for event A, 1/5 for

event B, and 2/7 for event C. Determine: (a) the probability of all three events happening; (b) the probability of event A and B but not C happening; (c) the probability of only event B happening; and (d) the probability of event A or event B happening but not

event C.

3. One bag contains 3 red and 5 black marbles and a second bag contains 4 green and 7 white marbles. One marble is drawn from the first bag and two marbles from the second bag, without replacement. Determine the probability of having;

(a) one red and two white marbles; (b) no green marbles; and (c) either one black and two green or one black and two white

marbles.


FEEDBACK TO ACTIVITY 1F

1. (a) 1/64 (f) 4096

3807

(b) 15/256

(c) 4096

2209 (g)

4096

735

(d) 256

75

(e) 39/64 2. (a) 1/140 (b) 1/156 (c) 1/8 (d) 13/56 3. (a) 63/440 (b) 21/55 (c) 27/88


1.7 CONDITIONAL PROBABILITY In the previous section, the events were independent of each other, since the occurrence of the first event in no way affected the outcome of the second event. On the other hand, when the occurrence of the first event changes the probability of the second event, the two events are said to be dependent . Refer to example 2 (b). The probability of selecting the blue ball is 10

2 . If the ball is not replaced,

then the probability of selecting the second (white) ball is 95 since there are only

nine balls remaining.

The conditional probability of an event B in relationship to event A is the probability that event B occurs after event A has already occurred. The notation for conditional probability is P(B/A). this notation does not mean that B is divided by A; rather, it means the probability that event B occurs given that event A has already occurred. In the ball example (ex 2b), P(B/A) is 9

2 since the first ball was not replaced.

When the outcome or occurrence of the first events affects the outcome or occurrence of the second event in such a way that the probability is changed, the events are said to be dependent.

INPUT


Multiplication Rule 2 When two events are dependent, the probability of both occurring is

P(A and B) =P(A).P(B/A) The conditional probability of an event B in relationship to an event A was defined as the probability that event B occurs after event A has already occurred. It can be found by dividing both sides of the equation for multiplication by P(A), as shown:

P(A and B) = P(A).P(B/A) therefore: )/()(

)(ABP

AP

AandBP =

The Venn diagram for conditional probability is shown in the figure below. In this case,

P(B/A) = )(

)&(

AP

BAP which is represented by the area in the

intersection or overlapping part of the circles A and B divided by area of circle A. The reasoning here is that if one assumes A has occurred, the A become the sample space for the next calculation and is the denominator of the probability

fraction)(

)&(

AP

BAP. The numerator P(A & B) represents the total probability of the

part B contained in A. Hence, P(A & B) becomes the numerator of the probability

fraction )(

)&(

AP

BAP. Imposing a condition reduces the sample space.


1.7.1 PROBABILITIES FOR “AT LEAST” The multiplication rules can be used with the complementary and non- complementary events to simplify solving probability problems involving “at least”. The next examples illustrate how this is done.

Venn diagram for Conditional probability

P(A) P(A and B) P(B)

P(B/A) = )(

)&(

AP

BAP


Example 1.7 1. In a shipment of 25 radios, 2 are defective. If two radios are randomly selected and tested, fond the probability that both are defective if the first one is not replaced after it has been tested. 2. Box 1 contains two red balls and one blue ball. Box 2 contains three blue balls and on red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. Find the probability of selecting a red ball. 3. A box contains white chips and black chips. A person selects two chips without replacement. If the probability of selecting a black chip and a white chip is 56

15 ,

and the probability of selecting a black chip on the first draw is 83 , find the

probability of selecting the white chip on the second draw, given that the first chip selected was a black chip. 4. The probability Ali parks in a no-parking zone and gets a parking ticket is 0.06, and the probability that Ali cannot find a legal parking space and has to park in the no-parking zone is 0.20. On Monday, Ali arrives at school and has to park in a no-parking zone. Find the probability that he will get a parking ticket. 5. A recent survey asked 100 people if they thought Sardin Cap Ayam is the best sardin. The results of the survey are shown in the table.

Gender Yes No Total Male 32 18 50 Female 8 42 50 Total 40 60 100

Find these probabilities. a. The respondent answered “yes,” given that the respondent was a female. b. The respondent was a male, given that the respondent answered “no.”


6. A coin is tossed five times. Find the probability of getting at least one tail. 7. A survey reported that 3% of pens sold in the Politeknik are Pilot pens. If 4

students who purchased a pen are randomly selected, find the probability that at least one purchased a Pilot pen.

8. A coin is tossed three times. Find the probability of getting i) Exactly 2 tails ii) At least 2 tails Solution to Example 1.7 1. Since the event are dependent, P(D1 and D2) =P(D1).P(D2/D1) = (2/25).(1/24) = 2/600 = 1/300 2. With the use of a tree diagram, the sample space can be determined as shown in the figure. First assign the probabilities to each branch. Next, using the multiplication rule, multiply the probabilities for each branch. Finally, use the addition rule, since a red ball can be obtained from box 1 or box2; P(red) = 24

11243

248

81

62 =+=+

Tree diagrams can be used when the events are independent or dependent, and they can also be used for sequences of three or more events.

Ball P(B1) ½ Box 1 P(R/B1) 2/3 Red 6

232

21 . =

P(B/B1) 1/3 Blue 6

131

21 . =

P(B2) ½ Box 2 P(R/B1) ¼ Red 8

141

21 . =

P(B/B2) ¾ Blue 8

343

21 . =


3. Let B = selecting a black chip W = selecting a white chip

Then P(W/B) = 7

5

)(

)&(

83

5615

==BP

WBP Hence the probability of selecting a white

chip on the second draw given that the first chip[ selected was black is 75 .

4. Let N= parking in a no-parking zone and T = getting a ticket, then

P(T/N) = 30.020.0

06.0

)(

)&( ==NP

TNP Hence, Ali has a 0.30 probability of

getting a parking ticket, given that he parked in a no-parking zone. 5. Let M = respondent was a male Y = respondent answered “yes” F = respondent was a female N = respondent answered “no” a) The problem is to find P(Y/F). The rules states P(Y/F) = P(Y and F) /P(F). The probability P(F and Y) is the number of females who responded “yes” divided by the total number of respondents: P(F and Y) =8/100 The probability P(F) is the probability of selecting a female: P(F) = 50/100 Then,

P(Y/F) = 25

4

100/50

100/8

)(

)&( ==FP

YFP

b. The problem is to find P(M/N)

P(M/N) = 10

3

100/60

100/18

)(

)&( ==NP

MNP


6. It is easier to find the probability of the complement of the event, which is

“all heads,” and then subtract the probability from 1 to get the probability of at least one tail.

P(E) = 1 – P(−E )

P(at least 1 tail) = 1 – P(all heads) P(all heads) = (1/2)5 = 32

1 Hence, P(at least one tail) = 1 - 32

31321 =

7. Let E = at least one Pilot pen purchased and −E = no Pilot pen purchased.

Then,

P(E) = 0.03 and P(−E ) = 1 – 0.03 = 0.97

P(no Pilot pen purchased) = (0.97)(0.97)(0.97) = 0.885; hence, P(at least one Pilot pen purchased) = 1 – 0.885 = 0.115

8. i) P(exactly 2 tails) = 8

381

81

81 =++

ii) P(at least 2 tails) = P(2 tails and 1 head) + P(3 tails) = 2

181

83 =+

(##Draw a tree diagram to verify your answer)


ACTIVITY 1G TEST YOUR UNDERSTANDING BEFORE YOU CONTINUE WITHY THE NEXT INPUT…!

1. The probabilities of an engine failing are given by: p1, failure due to overheating; p2, failure due to ignition problems; p3, failure due to fuel blockage. When p1 = 1/7, p2 = 2/9 and p3 = 3/11, determine the probabilities of:

(a) both p1 and p2 happening; (b) either p2 or p3 happening (c) both p1 and either p2 or p3 happening.

2. Actuarial tables show that the life expectancy of three men, A, B and C,

over a twenty-year period depends on their age and is given by PA = 4/15, PB = 11/15 and PC = 14/15. Determine the probabilities that in twenty years:

(a) all three men will be alive; (b) A will be alive but B and C will be dead; (c) At least one man will be alive.

3. When exploration for oil occurs a test hole is drilled. If as a result of this test drilling it seems likely that really large quantity of oil exist, (a bonanza) then the well is said to have structure. Examination of past records reveals the following information;

- Structure No structure Total Bonanza 0.20 0.05 0.25 No Bonanza 0.15 0.60 0.75 Total 0.35 0.65 1.00

a) Find P(Bonanza/structure) b) (No Bonanza/structure) c) P(Bonanza/no structure) d) P(No bonanza/no structure)


4. Suppose we have 100 urns. Type 1 urn (of which there are 70) each contains 5 black and 5 white balls. Type 2 urn (which there are 30) each contains 8 black and 2 white balls. An urn is randomly selected and a ball is drawn from that urn. If the ball chosen was black, what is the probability the ball came from a type 1 urn?


FEEDBACK TO ACTIVITY 1G 1. (a) 2/63 (b) 49/99 (c) 7/99

2. (a) 3375

616 (b)

3375

16 (c)

3375

3331

3. (a) 0.571 (b) 0.429 (c) 0.077 (d) 0.923 4. 0.593


SELF-ASSESMENT 1 You are approaching success. Try all the questions in this self assessment section and check your answers with those given in the Feedback to the Self -Assessment 1 on the next page. If you have problems, consult your instructor. Good luck. 1. State which events are independent and which are dependent.

i) Tossing a coin and throwing a die ii) Drawing a ball from an urn, not replacing it, and then drawing a second ball. iii) Getting a raise in salary and purchasing a new car. iv) Driving on ice and having accident. v) Having a large show size and having a high IQ. vi) A father being left-handed and a daughter being left-handed. vii) Smoking excessively and having lung cancer. viii) Eating an excessive amount of ice cream and smoking an excessive amount of cigarettes.

2. A survey found that 68% of book buyers are 40 or older. If two book buyers are selected at random, find the probability that both are 40 or older. 3. A salesman finds that the probability of making a sale is 0.23. If he talks to four customers today, find the probability that he will make four sales. 4. Find the probability of selecting two people at random who were born in the same month. 5. If three people are selected, find the probability that all three were born in January. 6. What is the probability that a husband, wife and daughter have the same birthday?


7. A radio uses six batteries, two of which are defective. If two are selected at random without replacement, find the probability that the first battery tests good and the second one is defective. 8. Out of 120 students, 90 of them put on white t-shirts. If five students are selected at random, one by one, find the probability that all will put on white t- shirts. 9. Urn 1 contains five red balls and three black balls. Urn 2 contains three red balls and one black ball. Urn 3 contains four red balls and two black balls. If an urn is selected at random and a ball is drawn, find the probability it will be red. 10. If a die is rolled three times, find the probability of getting at least one even number .


FEEDBACK TO SELF-ASSESMENT 1 Have you tried all the questions?? If “YES”, check your answer now. 1. a. Independent e. Independent b. Dependent f. Dependent c. Dependent g. dependent d. Dependent h. Independent 2. 0.462 3. 0.003 4. 1/12 5. 1/1728 6. 1/133,225 7. 4/15 8. 243/1024 9. 49/72 10. 7/8

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Note 2 probability

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