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Probability and Probability Distribution

Dr Manoj Kumar BhambuGCCBA-42, ChandigarhM- +91-988-823-7733

Probability and Probability Distribution: Definitions- Probability Rules –Application of Probability Rules-Conditional Probability- Bayes theorem

Random Variable and Probability Distributions; BinomialDistribution- Poisson Distribution and Normal Distribution.

Objectives of Learning this Unit

A probability is a quantitative measure of uncertainty – a number that conveys the strength of our belief in the occurrence of an uncertain event.

Probability is a measure of uncertainty of events in a random experiment.

Probability: Meaning

1. Deterministic Experiment: Experiments that have fixed outcome or result no matter any number of times they are repeated

2. Random Experiment: if an experiment, when repeated under identical conditions, do not produce the same outcome every time but the outcome in a trial is one of the several possible outcomes then such an experiment is known as random or probabilistic experiment.

Types of Experiment

A Random Experiment is any situation whose outcome cannot be predicted with certainty. Examples of an experiment include rolling a die, flipping a coin, and choosing a card from a deck of playing cards.

By an outcome or simple event we mean any result of the experiment. For example, the experiment of rolling a die yields six outcomes, namely, the outcomes 1,2,3,4,5, and 6.

Basic Terms in Probability

If you toss a coin you have an equal chance of getting a head or a tail. Heads or tails are the ‘outcomes’.

If a baby is born it has an equal chance of being a boy or a girl. Boy or girl are the ‘outcomes’.

If you roll a dice you have an equal chance of getting the number 1, 2, 3, 4, 5 or 6! These scores are the ‘outcomes’.

Examples of ‘Outcomes’

The sample space S of an experiment is the set of all possible outcomes for the experiment. For example, if you roll a die one time then the experiment is the roll of the die. A sample space for this experiment could be S = {1, 2, 3, 4, 5, 6} where each digit represents a face of the die.

Discrete Sample Space: A sample space whose elements are finite or infinite but countable is called Discrete Sample Space

Continuous Sample Space: A Sample Space whose elements are infinite and uncountable or assume all the values on a real line R or on an intervals of R is called a Continuous Sample Space.

Basic Terms in Probability (con.)

Event: An event is a subset of the sample space. For example, the event of rolling an odd number with a die consists of three simple events {1, 3, 5}.

Null Event: An event having no sample point is called a Null Event and is denoted by Φ.

Simple Event: An event consisting of only one sample point of a sample space is called a simple event.

Compound Events: When an event is decomposable into a number of simple events, then it is called a compound event.

Exhaustive Events: It is the total number of all possible outcomes of an experiment.

Mutually Exclusive Events: If in an experiment the occurrence of an event prevents or rules out the happening of all other events in the same experiment, then these events are said to be mutually exclusive events.

Equally Likely Events: The outcomes of an experiment are equally likely to occur when the probability of each outcome is equal.

Collectively Exhaustive Events : The total number of events in a population exhausts the population. So they are known as collectively exhaustive events.

Favourable cases: The cases which ensures the occurrence of an event are said to be favourable cases to the event.

When an experiment is conducted in such a way that the occurrence of event neither effect the occurrence of another event nor is effected by the occurrence of another event, both the events are called independent events.

Events which are not independent are called dependent events

Independent and Dependent Events

We can describe the probability or chance of an event happening by saying:

It is IMPOSSIBLE. It is UNLIKELY. It is LIKELY. It is CERTAIN.

Mathematical Vocabulary

If an experiment has n mutually exclusive, equally likely and exhaustive cases, out of which m are favourable to the happening of event A, then the probability of the happening of A is denoted by P(A) and is defined as:

P(A) = =

Probability of an event which is certain to occur is 1 and the probability of an event impossible is zero.

The probability of occurrence of any event lies between 0 and 1, both inclusive

Classical Definition of Probability

What is the probability of getting an even number in a single throw of a dice.

Answer: The possible cases in the throw of a dice are six viz. 1, 2, 3, 4, 5, 6.

Favourable cases are those which are marked with 2, 4, and 6.

∴ Probability of getting an even no = = =

Example 1

What is the probability of getting tail in a throw of a unbiased coin?

A bag contains 6 white balls and 9 black balls. What is the probability of drawing a black ball in a single draw?

What is the probability when a card is drawn at random from an ordinary pack of cards, if it is (i) a red card; (ii) a club; (iii) one of the court cards?

What is the probability of throwing a number greater than 3 with an ordinary dice?

What is the probability that a leap year selected at random will have 53 Sundays?

What is the probability of getting a total of more than 10 in a single throw of two dice?

More Examples

“ if the experiment be repeated a large number of times under essentially identical conditions, the limiting value of the ratio of the number of times the event A happens to the total number of trials of the experiments as the number of trials increases indefinitely is called the probability of happening of the event A.” -Von Mises

Symbolically, Let P(A) denote the probability of occurrence of event A, m be the number of times in which the event A occurs in a series of n trials then

P(A)= provided the limit is finite and unique

Statistical or Empirical definition of Probability

Definition. A probability function P on a finite sample space Ω assigns to each event A in Ω a number P(A) in [0,1] such that

(i) P(Ω) = 1, and(ii) P(A ∪ B) = P(A) + P(B) if A and B are

disjoint.The number P(A) is called the probability

that A occurs.

Modern Definition of Probability

Greek Alphabets

There are two important theorems of probability, namely

1. Addition Theorem or Theorem on Total Probability

2. Multiplication Theorem or Theorem on Compound Probability

Theorems on Probability

Statement: If n events are mutually exclusive, then the probability of happening of any one of them is equal to the sum of the probabilities of the happening of the separate events. In other words, if , …….. be n events and P(), P(), P(), …….P(), be their respective probabilities, thenP( ……..+ )=P((+……..+P( )

Addition Theorem

Proof: Let , …….. be mutually exclusive events with probabilities P(), P(), P(), …….P() respectively. Let N be the total number of trials. Let , …….. be the cases favourable to events , …….. respectively, then

P() =, i= 1, 2, ………, n.Since the events , …….. are mutually exclusive, therefore, the cases favourable to the happening of any one of the events , …….. are , …….. Hence the probability of happening of any one of the events , …….. isP( +……..+ ) = = + + ...... +

=P((+……..+P( )

A dice is rolled. What is the probability of getting a 1 or 6?

Answer: Let , be the event of getting a 1 and 6 respectively when a dice is rolled.

Here P()= P() = ∴ P(+ = P()P() ( By Addition Theorem)

= + = =

Example

If the probability of horse A winning the race is and the probability of horse B winning the race is , what is the probability that one of the horse will win the race?

Answer: Let , be the event of winning the race by horse A and horse B respectively.

Here P()= P() = ∴ P(+ = P()P() ( By Addition Theorem)

= + = =

Example

EXAMPLES ILLUSTRATING THE APPLICATION OF THE ADDITION THEOREM

Question: A card is drawn from a pack of 52 cards. What is the probability of getting either a king or a queen?

Solution: There are 4kings and 4 queens in a pack of 52 cards.The probability of drawing a king card is P (K) = and the probability of drawing a queen card is P (Q) =

Since, both the events are mutually exclusive, the probability that the card drawn is either a king or a queen isP (K or Q) = P (K) + P (Q) ( By Addition Theorem)

= + = =

Question :A card is drawn at random from a pack of cards. Find the probability that the drawn card is either a club or an ace of diamond.

Solution :Probability of drawing a club P (A) = Probability of drawing an ace of diamond P (B)

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

Theorem)

+ = =

More Example

Question: An investment consultant predicts that the odds against the price of a certain stock will go up during the next week are 2 : 1 and the odds in favour of the price remaining the same are 1 : 3. What is the probability that the price of the stock will go down during the next week.Solution: Let A denote the event ‘stock price will go up’, and B be the event ‘stock price will remain same’.Then P (A) = , and P (B) = P (stock price will either go up or remain same) = P (A U B)P (A) + P (B) = + = Now, P (stock price will go down) = 1 – P (A U B)

= 1 - =

ADDITION THEOREM FOR NOT MUTUALLY EXCLUSIVE EVENTS

Two or more events are known as partially overlapping if part of one event and part of another event occur together. Thus, when the events are not mutually exclusive the addition theorem has to be modified.Modified Addition Theorem states that if A and B are not mutually exclusive events, the probability of occurrence of either A or B or both is equal to the probability of that event A occurs, plus the probability that event B occurs minus the probability that events common to both A and B occur simultaneously. Symbolically,

P (A or B or Both ) = P (A) + P (B) – P (AB)

The following figure illustrates this point:

NOT MUTUALLY EXCLUSIVE EVENTS

Overlapping Events

A BAB

GeneralizationThe theorem can be extended to three or more events. If A, B and C are not mutually exclusive events, thenP (Either A or B or C) = P (A) + P (B) + P (C) – P (AB) – P (AC) – P (BC) + P (ABC)

EXAMPLES ILLUSTRATING THE APPLICATION OF THE MODIFIED ADDITION THEOREM

Question: A card is drawn at random from a well shuffled pack of cards. What is the probability that it is either a spade or a king ?

Solution: The Probability of drawing a spade P (A) =

The Probability of drawing a King P (B) =

The Probability of drawing a King of Spade P (AB) = P ( A or B or Both) = P (A) + P (B) – P (AB)

= + - = =

Question: A bag contains 30 balls number from 1 to 30. One ball is drawn at random. Find the probability that the number of ball is a multiple of 5 or 6.

Solution: Probability of the ball being a multiple of 5 P (A) =Probability of the ball being a multiple of 6 P (B) = Since, 30 is a multiple of 5 as well as 6, therefore the events are not mutually exclusive.P (A and B) = (common multiple 30)P (A or B) = P (A) + P (B) – P (AB) = + - =

MULTIPLICATION THEOREM: Independent EventsThe Multiplication Theorem states that if A and B are two

independent events, then the probability of the simultaneous occurrence of A and B is equal to the product of their individual probabilities. Symbolically,

P (AB) = P (A) X P (B)

Let be the number of cases favourable to the happening of the event A out of exhaustive and equally likely cases.

P (A) = Let be the number of cases favourable to the happening of the event B out of exhaustive and equally likely cases.

P (B) = Now, by the Fundamental Principle of counting , the number of cases favourable to the happening of the event AB is out of

P (AB) = P (AB) = P (A).P (B)

Hence, the theorem is proved.

GeneralisationThe theorem can be extended to three or more independent events. If A, B and C are three independent events, then

P (ABC) = P (A) x P (B) x P (C)

EXAMPLES ILLUSTRATING THE APPLICATION OF THE MULTIPLE THEOREM

Question: A coin is tossed three times. What is the probability of getting all the three heads?Solution: Probability of head in the first toss P (A) = Probability of head on the second toss P(B) =

Probability of head on the third toss P(C) = P (ABC) = P (A) x P (B) x P(C)

= x x =

Question: From a pack of 52 cards, two cards are drawn at random one after the another with replacement. What is the probability that both cards are kings?

Solution: The probability of drawing a King P (A) =

The probability of drawing again a king after replacement P (B) =

P (AB) = P (A) x P (B)

= x =

Probability of happening of atleast one event in case of n independent events

P (happening of at least one of the events) = 1 – P (happening of none of the events)

Question: A problem in statistics is given to three students. A, B and C whose chances od solving it are ½, 1/3 and ¼. What is the probability that the problems will be solved?Solution: Probability that A will solve the problem = P (A) = Probability that B will solve the problem = P (B) = Probability that C will solve the problem = P (C) = Probability that A will not solve the problem = P ( ) = 1 - = Probability that B will not solve the problem = P ( ) = 1 - = Probability that C will not solve the problem = P ( ) = 1 - = P (that none will solve the problem) = P ( ).P ( ).P ( )

= x x =

P (that problem will be solved) = 1 – P (that none will solve)

= 1 - =

Question: A candidate (Mr. X) is interviewed for 3 posts. For the first post, there are 3 candidates, for the second post, there are 4 and for the third there are 2. What are the chances of Mr. X being selected ?Solution : Probability of selection for 1st post = P (A) = Probability of selection for 2nd post = P (B) = Probability of selection for 3rd post = P (C) =

Probability of not getting selected for 1st post = P () = 1 - = Probability of not getting selected for 2nd post = P () = 1 - = Probability of not getting selected for 3rd post = P () = 1 - = P () = P ( ).P ( ).P ( )

= x x =

Probability of selection for at least 1 post = 1 – P (not selected at all)

= 1 - =

CONDITIONAL PROBABILITY

Dependent events are those in which the occurrence of one event affects the probability of other events. The probability of the event B given that A has occurred is called conditional probability of B. It is denoted by P (b/A). Similarly, the conditional probability of A given that B has occurred is denoted by P (A/B).

If A and B are two dependent events, then the conditional probability of B given A is defined and given by:

P (B/A) = provided P (A) > 0

Similarly, the conditional probability of A given B is defined and given by:

P (A/B) = provided P (B) > 0

MULTIPLICATION THEOREM IN CASE OF DPENDENT VARIABLES

When the events are not independent, i.e., they are dependent events, then the multiplication theorem has to be modified. The Modified Multiplication Theorem states that if A and B are two dependent events, then the probability of their simultaneous occurrence is equal to the probability of one event multiplied by the conditional probability of the other.

P (AB) = P (A) . P (B/A)OR

P (AB) = P (B) . P (A/B)

Where, P (B/A) = Conditional Probability of B given A.P (A/B) = Conditional Probability of A given B.

EXAMPLES ILLUSTRATING THE APPLICATION OF THE MODIFIED MULTIPLICATION THEOREM

Question: A bag contains 10 white and 5 black balls. Two balls are drawn at random one after the other without replacement. Find the probability that both balls drawn are black.Solution: The probability of drawing a black ball in the first attempt is: P (A) = = The probability of drawing the second black ball given that the first drawn is black and not replaced is: P (B/A) = = Since, the events are dependent, so the probability that both balls drawn are black is: x =

Question: Find the probability of drawing a king, a queen and a knave in that order from a pack of cards in three consecutive draws, the cards drawn not being replaced.Solution: The probability of drawing a king = P (A) = The probability of drawing a queen after a king has been drawn

P (B/A) = The probability of drawing a knave after a king and a queen have been drawn

P (C/AB) =

P (ABC) = x x =

COMBINED USE OF ADDITION AND MULTIPLICATION THEOREM

Question: A speaks truth in 80% cases, B in 90% cases. IN what percentage of cases are they likely to contradict each other in stating the same fact.Solution: Let P (A) and P (B) denote the probability that A and B speak the truth. Then,P (A) = = P () = 1 – P (A) = 1 - = P (B) = = P ( ) = 1 – P (B) = 1 - = They will contradict each other only when one of them speaks the truth and the other speaks a lie.Thus, there are two possibilities:(i) A Speaks the truth and B tells a lie(ii) B Speaks the truth and A tells a lie.P (A) = x = P ( B)= x =

Required Probability = + =

Question: A bag contains 5 white and 3 red balls and four balls are successively drawn out and not replaced. What is the chance that they are alternatively of different colours?Solution: 4 balls of alternative colours can be either White, Red, White, Red or Red, White, Red, WhiteBeginning with White Ball:The probability of drawing a white ball = The probability of drawing a red ball = The probability of drawing a white ball = The probability of drawing a red ball = P (1W 1R 1W 1R) = x x x =

Beginning with Red Ball: The probability of drawing a red ball = The probability of drawing a white ball = The probability of drawing a red ball = The probability of drawing a white ball = P (1R 1 W 1R 1W) = x x x =

Required Probability

= + =

A Permutation is an arrangement in a definite order of n objects taken r at a time.

nPr = n (n-1) (n-2)….(n-r+1)

= 0 < r ≤ n and n! = n (n-1) (n-2)………3.2.1Where repetition is not allowed.nPr = where repetition is allowed.

The number of permutation of n objects where p objects are of same kind and rest are different

= The number of permutation of n objects where objects are of one

kind, objects are of one kind….. objects are of one kind and rest if any are different

=

Permutation

Find the number of permutations of the letters of the word ALLHABAD.

Here Total no of Letters = 9A’s or = 4L’s or = 2

Required No. = = = = = 7560How many 4-digit numbers can be formed by using

the digits 1 to 9 if (1) repetition is allowed (2) repetition is not allowed.

when repetition is allowed nPr = = = 9999= 6561

when repetition is not allowed nPr = = = =3024

Find the number of permutations of the letters of the word ALLHABAD.

Here Total no of Letters = 9A’s or = 4L’s or = 2

Required No. = = = = = 7560How many 4-digit numbers can be formed by using

the digits 1 to 9 if (1) repetition is allowed (2) repetition is not allowed.

when repetition is allowed nPr = = = 9999= 6561

when repetition is not allowed nPr = = = =3024

A Combination is an arrangement of n objects taken r at a time.

nCr = n (n-1) (n-2)….(n-r+1)/ r!

= 0 < r ≤ n and n! = n (n-1) (n-2)………3.2.1

Note : Order is not important in case of combination

Combination

A committee of 3 persons is to be constituted from a group of 2 men and 3 women. What is the probability that the committee would consist of 1 man and two women.

Total Persons = 2+3 = 53 persons can be chosen out of 5 = 5C3 = = 10 ways1 man can be chosen out of 2 = 2C1 = = 2 ways

2 Women can be chosen out of 3 = 3C2 = = 3 ways

Required Probability = = =

Example

BERNOULLI’S THEOREM IN THE THEORY OF PROBABILITY

Bernoulli’s theorem is very useful in working out various probability problems. This theorem states that if the probability of happening of an event on one trial or experiment is known, then the probability of its happening exactly, 1,2,3,…r times in n trials can be determined by using the formula:

P (r) = nCr pr . qn-r r = 1,2,3,…n

Where,P (r) = Probability of r successes in n trials. p = Probability of success or happening of an event in one

trial. q = Probability of failure or not happening of the event in

one trial. n = Total number of trials.

A Box of 9 golf gloves contains 2 left-handed and 7 right handed. If 2 gloves are randomly selected that both gloves are of (i) of same hand (ii) of right hand (iii) of both hand.

(i) P(LL or RR) = + = + = = (ii) P(RR)= = = (iii) P (Both Hands) =( × ) + ( × ) =( × )+ (×)

=

Example

EXAMPLES ILLUSTRATING THE APPLICATION OF BERNOULLI’S THEOREM

Question: Three coins are tossed simultaneously. What is the probability that there will be exactly two heads?

Solution: P (r) = = nCrpr . qn-r

Given, n = 3, r = 2, p = probability of head in throw of a coin =

q = 1 - =

P (2H) = 3C2 () 2 . 3-2

= x = 3 x

=

BAYE’S THEOREM

Bayes' Theorem is named after the British Mathematician Thomas Bayes and it was published in the year 1763. With the help of Bayes' Theorem, prior probability are revised in the light of some sample information and posterior probabilities are obtained. This theorem is also called Theorem of Inverse Probability.

STATEMENT OF BAYE’S THEOREM

If A1 and A2 are mutually exclusive and exhaustive events and B be an event which can occur in combination with A1 and A2, then the conditional probability for event A1 and A2 given the event B is given by:P (A1/B) =

Similarly,P (A2/B) =

PROOF OF THE THEOREM

Since, A1 and A2 are mutually exclusive events and since the event B occurs with only one of them, so that

B = BA1 + BA2 or B = A1B + A2B

By the addition theorem of probability, we have

P (B) = P (A1B) + P(A2B) …(i)

A1A2

B

Now, by the multiplication theorem, we haveP (A1B) = P (A1) . P (B/A1) …(ii)P (A2B) = P (A2) . P (B/A2) …(iii)Substituting the values of P (A1B) and P (A2B) in (i), we getP (B) = P (A1) . P (B/A1) + P (A2) . P (B/A2) …(iv)Hence,P (B) = P (Ai) . P (B/Ai) …(v)Again by the theorem of conditional probability, we have

P (A1/B) = …(vi)Substituting the values of P (A1B) and P (B) from (ii) and (iv) in equation (vi), we get

P (A1/B) = Similarly,P (A2/B) =

The probabilities P (A1) and P (A2) are called posterior probabilities and probabilities P (A1/B) and P (A2/B) are called posterior probabilities.

The theorem can be expressed by means of the following figure:

P (B/A1) P (A1) . P (B/A1)

First Branch

Second Branch

P (A1)

P (A2)

P (B/A2) P (A2) . P (B/A2)

Prior Probability Conditional Probability Joint ProbabilityOf A1 and A2 of B given A1 and A2

Now, P (A1/B) =

=

Similarly,

P (A2/B) =

=

EXAMPLES ILLUSTRATING THE APPLICATION OF BAYE’S THEOREM

Question: In a bolt factory machine A, B and C manufacture respectively 25%, 35% and 40%

of the total. Of their output 5, 4, 2 per cent are defective bolts. A bolt is drawn at random from the product and is fount to be defective. What is the probability that it was manufactured by machine C?Solution: Let A, B and C be the events of drawing a bolt produced by machine A, B and C respectively and let D be the event that the bolt is defective.

The prior probabilities are: The conditional probabilities are:

P (A) = 25% = = 0.25 P (D/A) = 5% = = 0.05P (B) = 35% = = 0.35 P (D/B) = 4% = = 0.04

P (C) = 40% = = 0.40 P (D/C) = 2% = = 0.02

Events(1)

Prior Probability(2)

Conditional Probability (3)

Joint Probability(2) X (3)

A P(A) = 0.25 P (D/A) = 0.05 0.25x0.05

B P (B) = 0.35 P (D/B) = 0.04 0.35x0.04

C P (C) = 0.40 P (D/C) = 0.02 0.40x0.02

P (C/D) =

=

= =

= 0.2318 or 23.18%

ExampleAn aircraft emergency locator transmitter (ELT) is a device designed to transmit a signal in the case of a crash. The Altigauge Manufacturing Company makes 80% of the ELTs, the Bryant Company makes 15% of them, and the Chartair Company makes the other 5%. The ELTs made by Altigauge have a 4% rate of defects, the Bryant ELTs have a 6% rate of defects, and the Chartair ELTs have a 9% rate of defects (which helps to explain why Chartair has the lowest market share).a. If an ELT is randomly selected from the general population of all ELTs, find the probability that it was made by the Altigauge Manufacturing Company.b. If a randomly selected ELT is then tested and is found to be defective, find the probability that it was made by the Altigauge Manufacturing Company.

A factory has two machines A and B. Past records show that machine A produces 30% of the total output and machine B 70%. Machine A produces 5% defective articles and machine B produces 1% defective items. An item is drawn at random and found to be defective. What is the probability that it was produced by machine A.

A man is known to speak truth 3 out of 4 times. He throws a dice and reports that it is a six. Find the probability that it is actually six.

A restaurant serves two special dishes, A and the rest B to its customers consisting of 60 % men and 40% women. 80% of men order dish A and the rest dish B. 70% of women order dish B and the rest A. In what ratio of A to B should the restaurant prepare the two dishes.

There are two identical boxes containing respectively 4 white and 3 red balls; and 3 white and 7 red balls respectively. A ball is drawn and it is white what is the probability that is drawn from first box.

More Examples

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