PROBABILITY

The theory of probabilities is simply the science of logic quantitatively treated. – C.S. PEIRCE 

13.1 Introduction

In earlier Classes, we have studied the probability as a measure of uncertainty of events in a random experiment. We discussed the axiomatic approach formulated by Russian Mathematician, A.N. Kolmogorov (1903-1987) and treated probability as a function of outcomes of the experiment. We have also established equivalence between the axiomatic theory and the classical theory of probability in case of equally likely outcomes. On the basis of this relationship, we obtained probabilities of events associated with discrete sample spaces. We have also studied the addition rule of probability. In this chapter, we shall discuss the important concept of conditional probability of an event given that another event has occurred, which will be helpful in understanding the Bayes' theorem, multiplication rule of probability and independence of events. We shall also learn an important concept of random variable and its probability distribution and also the mean and variance of a probability distribution. In the last section of the chapter, we shall study an important discrete probability distribution called Binomial distribution. Throughout this chapter, we shall take up the experiments having equally likely outcomes, unless stated otherwise.

13.2 Conditional Probability

Uptill now in probability, we have discussed the methods of finding the probability of events. If we have two events from the same sample space, does the information about the occurrence of one of the events affect the probability of the other event? Let us try to answer this question by taking up a random experiment in which the outcomes are equally likely to occur.

Consider the experiment of tossing three fair coins. The sample space of the experiment is

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Since the coins are fair, we can assign the probability to each sample point. Let E be the event ‘at least two heads appear’ and F be the event ‘first coin shows tail’. Then

E = {HHH, HHT, HTH, THH}

and F = {THH, THT, TTH, TTT}

Therefore P(E) = P ({HHH}) + P ({HHT}) + P ({HTH}) + P ({THH})

= (Why ?)

and P(F) = P ({THH}) + P ({THT}) + P ({TTH}) + P ({TTT})

=

Also E ∩ F = {THH}

with P(E ∩ F) = P({THH}) =

Now, suppose we are given that the first coin shows tail, i.e. F occurs, then what is the probability of occurrence of E? With the information of occurrence of F, we are sure that the cases in which first coin does not result into a tail should not be considered while finding the probability of E. This information reduces our sample space from the set S to its subset F for the event E. In other words, the additional information really amounts to telling us that the situation may be considered as being that of a new random experiment for which the sample space consists of all those outcomes only which are favourable to the occurrence of the event F.

Now, the sample point of F which is favourable to event E is THH.

Thus, Probability of E considering F as the sample space = ,

or Probability of E given that the event F has occurred =

This probability of the event E is called the conditional probability of E given that F has already occurred, and is denoted by P (E|F).

Thus P(E|F) =

Note that the elements of F which favour the event E are the common elements of E and F, i.e. the sample points of E ∩ F.

Thus, we can also write the conditional probability of E given that F has occurred as

P(E|F) =

=

Dividing the numerator and the denominator by total number of elementary events of the sample space, we see that P(E|F) can also be written as

P(E|F) = ... (1)

Note that (1) is valid only when P(F) ≠ 0 i.e., F ≠ φ (Why?)

Thus, we can define the conditional probability as follows :

Definition 1 If E and F are two events associated with the same sample space of a random experiment, the conditional probability of the event E given that F has occurred, i.e. P (E|F) is given by

P(E|F) = provided P(F) ≠ 0

13.2.1 Properties of conditional probability

Let E and F be events of a sample space S of an experiment, then we have

Property 1 P(S|F) = P(F|F) = 1

We know that

P(S|F) =

Also P(F|F) =

Thus P(S|F) = P(F|F) = 1

Property 2 If A and B are any two events of a sample space S and F is an event of S such that P(F) ≠ 0, then

P((A ∪ B)|F) = P(A|F) + P(B|F) – P((A ∩ B)|F)

In particular, if A and B are disjoint events, then

P((A∪B)|F) = P(A|F) + P(B|F)

We have

P((A∪B)|F) =

=

(by distributive law of union of sets over intersection)

=

=

= P(A|F) + P(B|F) – P((A∩B)|F)

When A and B are disjoint events, then

P((A ∩ B)|F) = 0

⇒ P((A ∪ B)|F) = P(A|F) + P(B|F)

Property 3 P(E′|F) = 1 − P(E|F)

From Property 1, we know that P(S|F) = 1

⇒ P(E ∪ E′|F) = 1 since S = E ∪ E′

⇒ P(E|F) + P (E′|F) = 1 since E and E′ are disjoint events

Thus, P(E′|F) = 1 − P(E|F)

Let us now take up some examples.

Example 1 If P(A) = , P(B) = and P(A ∩ B) = , evaluate P(A|B).

Solution We have

Example 2 A family has two children. What is the probability that both the children are boys given that at least one of them is a boy ?

Solution Let b stand for boy and g for girl. The sample space of the experiment is

S = {(b, b), (g, b), (b, g), (g, g)}

Let E and F denote the following events :

E : ‘both the children are boys’

F : ‘at least one of the child is a boy’

Then E = {(b,b)} and F = {(b,b), (g,b), (b,g)}

Now E ∩ F = {(b,b)}

Thus P(F) = and P (E ∩ F )=

Therefore P(E|F) =

Example 3 Ten cards numbered 1 to 10 are placed in a box, 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?

Solution Let A be the event ‘the number on the card drawn is even’ and B be the event ‘the number on the card drawn is greater than 3’. We have to find 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) =

Then P(A|B) =

Example 4 In a school, there are 1000 students, out of which 430 are girls. It is known that out of 430, 10% of the girls study in class XII. What is the probability that a student chosen randomly studies in Class XII given that the chosen student is a girl?

Solution Let E denote the event that a student chosen randomly studies in Class XII and F be the event that the randomly chosen student is a girl. We have to find P (E|F).

Now P(F) = and (Why?)

Then P(E|F) =

Example 5 A die is thrown three times. Events A and B are defined as below:

A : 4 on the third throw

B : 6 on the first and 5 on the second throw

Find the probability of A given that B has already occurred.

Solution The sample space has 216 outcomes.

Now A =

B = {(6,5,1), (6,5,2), (6,5,3), (6,5,4), (6,5,5), (6,5,6)}

and A ∩ B = {(6,5,4)}.

Now P(B) = and P (A ∩ B) =

Then P(A|B) =

Example 6 A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once?

Solution Let E be the event that ‘number 4 appears at least once’ and F be the event that ‘the sum of the numbers appearing is 6’.

Then, E = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (1,4), (2,4), (3,4), (5,4), (6,4)}

and F = {(1,5), (2,4), (3,3), (4,2), (5,1)}

We have P(E) = and P(F) =

Also E∩F = {(2,4), (4,2)}

Therefore P(E∩F) =

Hence, the required probability

P(E|F) =

For the conditional probability discussed above, we have considered the elementary events of the experiment to be equally likely and the corresponding definition of the probability of an event was used. However, the same definition can also be used in the general case where the elementary events of the sample space are not equally likely, the probabilities P(E∩F) and P(F) being calculated accordingly. Let us take up the following example.

Example 7 Consider the experiment of tossing a coin. If the coin shows head, toss it again but if it shows tail, then throw a die. Find the conditional probability of the event that ‘the die shows a number greater than 4’ given that ‘there is at least one tail’.

Solution The outcomes of the experiment can be represented in following diagrammatic manner called the ‘tree diagram’.

The sample space of the experiment may be described as

S = {(H,H), (H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}

where (H, H) denotes that both the tosses result into head and (T, i) denote the first toss result into a tail and the number i appeared on the die for i = 1,2,3,4,5,6.

Thus, the probabilities assigned to the 8 elementary events

(H, H), (H, T), (T, 1), (T, 2), (T, 3) (T, 4), (T, 5), (T, 6) are respectively which is clear from the Fig 13.2.

Let F be the event that ‘there is at least one tail’ and E be the event ‘the die shows a number greater than 4’. Then

F = {(H,T), (T,1), (T,2), (T,3), (T,4), (T,5), (T,6)}

E = {(T,5), (T,6)} and E ∩ F = {(T,5), (T,6)}

Now P(F) = P({(H,T)}) + P ({(T,1)}) + P ({(T,2)}) + P ({(T,3)})

+ P ({(T,4)}) + P({(T,5)}) + P({(T,6)})

=

and P(E ∩ F) = P ({(T,5)}) + P ({(T,6)}) =

Hence P(E|F) =

13.3 Multiplication Theorem on Probability

Let E and F be two events associated with a sample space S. Clearly, the set E ∩ F denotes the event that both E and F have occurred. In other words, E ∩ F denotes the simultaneous occurrence of the events E and F. The event E ∩ F is also written as EF.

Very often we need to find the probability of the event EF. For example, in the experiment of drawing two cards one after the other, we may be interested in finding the probability of the event ‘a king and a queen’. The probability of event EF is obtained by using the conditional probability as obtained below :

We know that the conditional probability of event E given that F has occurred is denoted by P(E|F) and is given by

P(E|F) =

From this result, we can write

P(E ∩ F) = P(F) . P(E|F) ... (1)

Also, we know that

P(F|E) =

or P(F|E) = (since E ∩ F = F ∩ E)

Thus, P(E ∩ F) = P(E). P(F|E) .... (2)

Combining (1) and (2), we find that

P(E ∩ F) = P(E) P(F|E)

= P(F) P(E|F) provided P(E) ≠ 0 and P(F) ≠ 0.

The above result is known as the multiplication rule of probability.

Let us now take up an example.

Example 8 An urn contains 10 black and 5 white balls. Two balls are drawn from the urn one after the other without replacement. What is the probability that both drawn balls are black?

Solution Let E and F denote respectively the events that first and second ball drawn are black. We have to find P(E ∩ F) or P(EF).

Now P(E) = P (black ball in first draw) =

Also given that the first ball drawn is black, i.e., event E has occurred, now there are 9 black balls and five white balls left in the urn. Therefore, the probability that the second ball drawn is black, given that the ball in the first draw is black, is nothing but the conditional probability of F given that E has occurred.

i.e. P(F|E) =

By multiplication rule of probability, we have

P(E ∩ F) = P(E) P(F|E)

=

Multiplication rule of probability for more than two events If E, F and G are three events of sample space, we have

P(E ∩ F ∩ G) = P(E) P(F|E) P(G|(E ∩ F)) = P(E) P(F|E) P(G|EF)

Similarly, the multiplication rule of probability can be extended for four or
more events.

The following example illustrates the extension of multiplication rule of probability for three events.

Example 9 Three cards are drawn successively, without replacement from a pack of 52 well shuffled cards. What is the probability that first two cards are kings and the third card drawn is an ace?

Solution Let K denote the event that the card drawn is king and A be the event that the card drawn is an ace. Clearly, we have to find P (KKA)

Now P(K) =

Also, P(K|K) is the probability of second king with the condition that one king has already been drawn. Now there are three kings in (52 − 1) = 51 cards.

Therefore P(K|K) =

Lastly, P(A|KK) is the probability of third drawn card to be an ace, with the condition that two kings have already been drawn. Now there are four aces in left 50 cards.

Therefore P(A|KK) =

By multiplication law of probability, we have

P(KKA) = P(K) P(K|K) P(A|KK)

=

13.4 Independent Events

Consider the experiment of drawing a card from a deck of 52 playing cards, in which the elementary events are assumed to be equally likely. If E and F denote the events 'the card drawn is a spade' and 'the card drawn is an ace' respectively, then

P(E) =

Also E and F is the event ' the card drawn is the ace of spades' so that

P(E∩F) =

Hence P(E|F) =

Since P(E) = = P (E|F), we can say that the occurrence of event F has not affected the probability of occurrence of the event E.

We also have

P(F|E) =

Again, P(F) = = P(F|E) shows that occurrence of event E has not affected the probability of occurrence of the event F.

Thus, E and F are two events such that the probability of occurrence of one of them is not affected by occurrence of the other.

Such events are called independent events.

Definition 2 Two events E and F are said to be independent, if

P(F|E) = P (F) provided P (E) ≠ 0

and P(E|F) = P (E) provided P (F) ≠ 0

Thus, in this definition we need to have P (E) ≠ 0 and P(F) ≠ 0

Now, by the multiplication rule of probability, we have

P(E ∩ F) = P(E) . P (F|E) ... (1)

If E and F are independent, then (1) becomes

P(E ∩ F) = P(E) . P(F) ... (2)

Thus, using (2), the independence of two events is also defined as follows:

Definition 3 Let E and F be two events associated with the same random experiment, then E and F are said to be independent if

P(E ∩ F) = P(E) . P (F)

Remarks

(i) Two events E and F are said to be dependent if they are not independent, i.e. if

P(E ∩ F ) ≠ P(E) . P (F)

(ii) Sometimes there is a confusion between independent events and mutually exclusive events. Term ‘independent’ is defined in terms of ‘probability of events’ whereas mutually exclusive is defined in term of events (subset of sample space). Moreover, mutually exclusive events never have an outcome common, but independent events, may have common outcome. Clearly, ‘independent’ and ‘mutually exclusive’ do not have the same meaning.

In other words, two independent events having nonzero probabilities of occurrence can not be mutually exclusive, and conversely, i.e. two mutually exclusive events having nonzero probabilities of occurrence can not be independent.

(iii) Two experiments are said to be independent if for every pair of events E and F, where E is associated with the first experiment and F with the second experiment, the probability of the simultaneous occurrence of the events E and F when the two experiments are performed is the product of P(E) and P(F) calculated separately on the basis of two experiments, i.e., P (E ∩ F) = P (E) . P(F)

(iv) Three events A, B and C are said to be mutually independent, if

P(A ∩ B) = P(A) P(B)

P(A ∩ C) = P(A) P(C)

P(B ∩ C) = P(B) P(C)

and P(A ∩ B ∩ C) = P(A) P(B) P(C)

If at least one of the above is not true for three given events, we say that the events are not independent.

Example 10 A die is thrown. If E is the event ‘the number appearing is a multiple of 3’ and F be the event ‘the number appearing is even’ then find whether E and F are independent ?

Solution We know that the sample space is S = {1, 2, 3, 4, 5, 6}

Now E = { 3, 6}, F = { 2, 4, 6} and E ∩ F = {6}

Then P(E) =

Clearly P(E ∩ F) = P(E). P (F)

Hence E and F are independent events.

Example 11 An unbiased die is thrown twice. Let the event A be ‘odd number on the first throw’ and B the event ‘odd number on the second throw’. Check the independence of the events A and B.

Solution If all the 36 elementary events of the experiment are considered to be equally likely, we have

P(A) = and

Also P(A ∩ B) = P (odd number on both throws)

=

Now P(A) P(B) =

Clearly P(A ∩ B) = P(A) × P(B)

Thus, A and B are independent events

Example 12 Three coins are tossed simultaneously. Consider the event E ‘three heads or three tails’, F ‘at least two heads’ and G ‘at most two heads’. Of the pairs (E,F), (E,G) and (F,G), which are independent? which are dependent?

Solution The sample space of the experiment is given by

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Clearly E = {HHH, TTT}, F= {HHH, HHT, HTH, THH}

and G = {HHT, HTH, THH, HTT, THT, TTH, TTT}

Also E ∩ F = {HHH}, E ∩ G = {TTT}, F ∩ G = { HHT, HTH, THH}

Therefore P(E) =

and P(E∩F) =

Also P(E) . P(F) =

and P(F) . P(G) =

Thus P(E ∩ F) = P(E) . P(F)

P(E ∩ G) ≠ P(E) . P(G)

and P(F ∩ G) ≠ P (F) . P(G)

Hence, the events (E and F) are independent, and the events (E and G) and (F and G) are dependent.

Example 13 Prove that if E and F are independent events, then so are the events E and F′.

Solution Since E and F are independent, we have

P(E ∩ F) = P(E) . P(F) ....(1)

From the venn diagram in Fig 13.3, it is clear that E ∩ F and E ∩ F′ are mutually exclusive events and also E = (E ∩ F) ∪ (E ∩ F′).

Therefore P(E) = P(E ∩ F) + P(E ∩ F′)

or P(E ∩ F′) = P(E) − P(E ∩ F)

= P(E) − P(E) . P(F) (by (1))

= P(E) (1−P(F))

= P(E). P(F′)

Hence, E and F′ are independent

Example 14 If A and B are two independent events, then the probability of occurrence of at least one of A and B is given by 1– P(A′) P(B′)

Solution We have

P(at least one of A and B) = P(A ∪ B)

= P(A) + P(B) − P(A ∩ B)

= P(A) + P(B) − P(A) P(B)

= P(A) + P(B) [1−P(A)]

= P(A) + P(B). P(A′)

= 1− P(A′) + P(B) P(A′)

= 1− P(A′) [1− P(B)]

= 1− P(A′) P (B′)

13.5 Bayes' Theorem

Consider that there are two bags I and II. Bag I contains 2 white and 3 red balls and Bag II contains 4 white and 5 red balls. One ball is drawn at random from one of the bags. We can find the probability of selecting any of the bags (i.e.) or probability of drawing a ball of a particular colour (say white) from a particular bag (say Bag I). In other words, we can find the probability that the ball drawn is of a particular colour, if we are given the bag from which the ball is drawn. But, can we find the probability that the ball drawn is from a particular bag (say Bag II), if the colour of the ball drawn is given? Here, we have to find the reverse probability of Bag II to be selected when an event occurred after it is known. Famous mathematician, John Bayes' solved the problem of finding reverse probability by using conditional probability. The formula developed by him is known as ‘Bayes theorem’ which was published posthumously in 1763. Before stating and proving the Bayes' theorem, let us first take up a definition and some preliminary results.

13.5.1 Partition of a sample space

A set of events E1, E2, ..., En is said to represent a partition of the sample space S if

(a) Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, 3, ..., n

(b) E1 ∪ Ε2 ∪ ... ∪ En= S and

(c) P(Ei) > 0 for all i = 1, 2, ..., n.

In other words, the events E1, E2, ..., En represent a partition of the sample space S if they are pairwise disjoint, exhaustive and have nonzero probabilities.

As an example, we see that any nonempty event E and its complement E′ form a partition of the sample space S since they satisfy E ∩ E′ = φ and E ∪ E′ = S.

From the Venn diagram in Fig 13.3, one can easily observe that if E and F are any two events associated with a sample space S, then the set {E ∩ F′, E ∩ F, E′ ∩ F, E′ ∩ F′} is a partition of the sample space S. It may be mentioned that the partition of a sample space is not unique. There can be several partitions of the same sample space.

We shall now prove a theorem known as Theorem of total probability.

13.5.2 Theorem of total probability

Let {E1, E2,...,En} be a partition of the sample space S, and suppose that each of the events E1, E2,..., En has nonzero probability of occurrence. Let A be any event associated with S, then

P(A) = P(E1) P(A|E1) + P(E2) P(A|E2) + ... + P(En) P(A|En)

=

Proof Given that E1, E2,..., En is a partition of the sample space S (Fig 13.4). Therefore,

S = E1 ∪ E2 ∪ ... ∪ En ... (1)

and Ei ∩ Ej = φ, i ≠ j, i, j = 1, 2, ..., n

Now, we know that for any event A,

A = A ∩ S

= A ∩ (E1 ∪ E2 ∪ ... ∪ En)

= (A ∩ E1) ∪ (A ∩ E2) ∪ ...∪ (A ∩ En)

Also A ∩ Ei and A ∩ Ej are respectively the subsets of Ei and Ej. We know that Ei and Ej are disjoint, for , therefore, A ∩ Ei and A ∩ Ej are also disjoint for all i ≠ j, i, j = 1, 2, ..., n.

Thus, P(A) = P [(A ∩ E1) ∪ (A ∩ E2)∪ .....∪ (A ∩ En)]

= P (A ∩ E1) + P (A ∩ E2) + ... + P (A ∩ En)

Now, by multiplication rule of probability, we have

P(A ∩ Ei) = P(Ei) P(A|Ei) as P (Ei) ≠ 0∀i = 1,2,..., n

Therefore, P (A) = P (E1) P (A|E1) + P (E2) P (A|E2) + ... + P (En)P(A|En)

or P(A) =

Example 15 A person has undertaken a construction job. The probabilities are 0.65 that there will be strike, 0.80 that the construction job will be completed on time if there is no strike, and 0.32 that the construction job will be completed on time if there is a strike. Determine the probability that the construction job will be completed on time.

Solution Let A be the event that the construction job will be completed on time, and B be the event that there will be a strike. We have to find P(A).

We have

P(B) = 0.65, P(no strike) = P(B′) = 1 − P(B) = 1 − 0.65 = 0.35

P(A|B) = 0.32, P(A|B′) = 0.80

Since events B and B′ form a partition of the sample space S, therefore, by theorem on total probability, we have

P(A) = P(B) P(A|B) + P(B′) P(A|B′)

= 0.65 × 0.32 + 0.35 × 0.8

= 0.208 + 0.28 = 0.488

Thus, the probability that the construction job will be completed in time is 0.488.

We shall now state and prove the Bayes' theorem.

Bayes’ Theorem  If E1, E2 ,..., En are n non empty events which constitute a partition of sample space S, i.e. E1, E2 ,..., En are pairwise disjoint and E1∪ E2∪ ... ∪ En = S and A is any event of nonzero probability, then

P(Ei|A) = for any i = 1, 2, 3, ..., n

Proof By formula of conditional probability, we know that

P(Ei|A) =

= (by multiplication rule of probability)

= (by the result of theorem of total probability)

Remark The following terminology is generally used when Bayes' theorem is applied.

The events E1, E2, ..., En are called hypotheses.

The probability P(Ei) is called the priori probability of the hypothesis Ei

The conditional probability P(Ei |A) is called a posteriori probability of the hypothesis Ei.

Bayes' theorem is also called the formula for the probability of "causes". Since the Ei's are a partition of the sample space S, one and only one of the events Ei occurs (i.e. one of the events Ei must occur and only one can occur). Hence, the above formula gives us the probability of a particular Ei (i.e. a "Cause"), given that the event A has occurred.

The Bayes' theorem has its applications in variety of situations, few of which are illustrated in following examples.

Example 16 Bag I contains 3 red and 4 black balls while another Bag II contains 5 red and 6 black balls. One ball is drawn at random from one of the bags and it is found to be red. Find the probability that it was drawn from Bag II.

Solution Let E1 be the event of choosing the bag I, E2 the event of choosing the bag II and A be the event of drawing a red ball.

Then P(E1) = P(E2) =

Also P(A|E1) = P(drawing a red ball from Bag I) =

and P(A|E2) = P(drawing a red ball from Bag II) =

Now, the probability of drawing a ball from Bag II, being given that it is red,
is P(E2|A)

By using Bayes' theorem, we have

P(E2|A) = =

Example 17 Given three identical boxes I, II and III, each containing two coins. In box I, both coins are gold coins, in box II, both are silver coins and in the box III, there is one gold and one silver coin. A person chooses a box at random and takes out a coin. If the coin is of gold, what is the probability that the other coin in the box is also of gold?

Solution Let E1, E2 and E3 be the events that boxes I, II and III are chosen, respectively.

Then P(E1) = P(E2) = P(E3) =

Also, let A be the event that ‘the coin drawn is of gold’

Then P(A|E1) = P(a gold coin from bag I) = = 1

P(A|E2) = P(a gold coin from bag II) = 0

P(A|E3) = P(a gold coin from bag III) =

Now, the probability that the other coin in the box is of gold

= the probability that gold coin is drawn from the box I.

= P(E1|A)

By Bayes' theorem, we know that

P(E1|A) =

=

Example 18 Suppose that the reliability of a HIV test is specified as follows:

Of people having HIV, 90% of the test detect the disease but 10% go undetected. Of people free of HIV, 99% of the test are judged HIV–ive but 1% are diagnosed as showing HIV+ive. From a large population of which only 0.1% have HIV, one person is selected at random, given the HIV test, and the pathologist reports him/her as HIV+ive. What is the probability that the person actually has HIV?

Solution Let E denote the event that the person selected is actually having HIV and A the event that the person's HIV test is diagnosed as +ive. We need to find P(E|A).

Also E′ denotes the event that the person selected is actually not having HIV.

Clearly, {E, E′} is a partition of the sample space of all people in the population. We are given that

P(E) = 0.1%

P(E′) = 1 – P(E) = 0.999

P(A|E) = P(Person tested as HIV+ive given that he/she is actually having HIV)

= 90%

and P(A|E′) = P(Person tested as HIV +ive given that he/she is actually not having HIV)

= 1% = = 0.01

Now, by Bayes' theorem

P(E|A) =

=

= 0.083 approx.

Thus, the probability that a person selected at random is actually having HIV given that he/she is tested HIV+ive is 0.083.

Example 19 In a factory which manufactures bolts, machines A, B and C manufacture respectively 25%, 35% and 40% of the bolts. Of their outputs, 5, 4 and 2 percent are respectively defective bolts. A bolt is drawn at random from the product and is found to be defective. What is the probability that it is manufactured by the machine B?

Solution Let events B1, B2, B3 be the following :

B1 : the bolt is manufactured by machine A

B2 : the bolt is manufactured by machine B

B3 : the bolt is manufactured by machine C

Clearly, B1, B2, B3 are mutually exclusive and exhaustive events and hence, they represent a partition of the sample space.

Let the event E be ‘the bolt is defective’.

The event E occurs with B1 or with B2 or with B3. Given that,

P(B1) = 25% = 0.25, P (B2) = 0.35 and P(B3) = 0.40

Again P(E|B1) = Probability that the bolt drawn is defective given that it is manufactured by machine A = 5% = 0.05

Similarly, P(E|B2) = 0.04, P(E|B3) = 0.02.

Hence, by Bayes' Theorem, we have

P(B2|E) =

=

=

Example 20 A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by other means of transport are respectively . The probabilities that he will be late are ,
if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train?

Solution Let E be the event that the doctor visits the patient late and let T1, T2, T3, T4 be the events that the doctor comes by train, bus, scooter, and other means of transport respectively.

Then P(T1) = (given)

P(E|T1) = Probability that the doctor arriving late comes by train =

Similarly, P(E|T2) = , P(E|T3) = and P(E|T4) = 0, since he is not late if he comes by other means of transport.

Therefore, by Bayes' Theorem, we have

P(T1|E) = Probability that the doctor arriving late comes by train

=

= =

Hence, the required probability is .

Example 21 A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

Solution Let E be the event that the man reports that six occurs in the throwing of the die and let S1 be the event that six occurs and S2 be the event that six does not occur.

Then P(S1) = Probability that six occurs =

P(S2) = Probability that six does not occur =

P(E|S1) = Probability that the man reports that six occurs when six has actually occurred on the die

= Probability that the man speaks the truth =

P(E|S2) = Probability that the man reports that six occurs when six has not actually occurred on the die

= Probability that the man does not speak the truth

Thus, by Bayes' theorem, we get

P(S1|E) = Probability that the report of the man that six has occurred is actually a six

=

=

Hence, the required probability is

13.6 Random Variables and its Probability Distributions

We have already learnt about random experiments and formation of sample spaces. In most of these experiments, we were not only interested in the particular outcome that occurs but rather in some number associated with that outcomes as shown in following examples/experiments.

(i) In tossing two dice, we may be interested in the sum of the numbers on the two dice.

(ii) In tossing a coin 50 times, we may want the number of heads obtained.

(iii) In the experiment of taking out four articles (one after the other) at random from a lot of 20 articles in which 6 are defective, we want to know the number of defectives in the sample of four and not in the particular sequence of defective and nondefective articles.

In all of the above experiments, we have a rule which assigns to each outcome of the experiment a single real number. This single real number may vary with different outcomes of the experiment. Hence, it is a variable. Also its value depends upon the outcome of a random experiment and, hence, is called random variable. A random variable is usually denoted by X.

If you recall the definition of a function, you will realise that the random variable X is really speaking a function whose domain is the set of outcomes (or sample space) of a random experiment. A random variable can take any real value, therefore, its
co-domain is the set of real numbers. Hence, a random variable can be defined as follows :

Definition 4 A random variable is a real valued function whose domain is the sample space of a random experiment.

For example, let us consider the experiment of tossing a coin two times in succession.

The sample space of the experiment is S = {HH, HT, TH, TT}.

If X denotes the number of heads obtained, then X is a random variable and for each outcome, its value is as given below :

X(HH) = 2, X (HT) = 1, X (TH) = 1, X (TT) = 0.

More than one random variables can be defined on the same sample space. For example, let Y denote the number of heads minus the number of tails for each outcome of the above sample space S.

Then Y(HH) = 2, Y (HT) = 0, Y (TH) = 0, Y (TT) = – 2.

Thus, X and Y are two different random variables defined on the same sample space S.

Example 22 A person plays a game of tossing a coin thrice. For each head, he is given Rs 2 by the organiser of the game and for each tail, he has to give Rs 1.50 to the organiser. Let X denote the amount gained or lost by the person. Show that X is a random variable and exhibit it as a function on the sample space of the experiment.

Solution X is a number whose values are defined on the outcomes of a random experiment. Therefore, X is a random variable.

Now, sample space of the experiment is

S = {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}

Then X(HHH) = Rs (2 × 3) = Rs 6

X(HHT) = X(HTH) = X(THH) = Rs (2 × 2 − 1 × 1.50) = Rs 2.50

X(HTT) = X(THT) = (TTH) = Rs (1 × 2) – (2 × 1.50) = – Re 1

and X(TTT) = − Rs (3 × 1.50) = − Rs 4.50

where, minus sign shows the loss to the player. Thus, for each element of the sample space, X takes a unique value, hence, X is a function on the sample space whose range is

{–1, 2.50, – 4.50, 6}

Example 23 A bag contains 2 white and 1 red balls. One ball is drawn at random and then put back in the box after noting its colour. The process is repeated again. If X denotes the number of red balls recorded in the two draws, describe X.

Solution Let the balls in the bag be denoted by w1, w2, r. Then the sample space is

S = {w1 w1, w1 w2, w2 w2, w2 w1, w1 r, w2 r, r w1, r w2, r r}

Now, for ω ∈ S

X(ω) = number of red balls

Therefore

X({w1 w1}) = X({w1 w2}) = X({w2 w2}) = X({w2 w1}) = 0

X({w1 r}) = X({w2 r}) = X({r w1}) = X({r w2}) = 1 and X({r r}) = 2

Thus, X is a random variable which can take values 0, 1 or 2.

13.6.1 Probability distribution of a random variable

Let us look at the experiment of selecting one family out of ten families f1, f2 ,..., f10 in such a manner that each family is equally likely to be selected. Let the families f1, f2,
... , f10 have 3, 4, 3, 2, 5, 4, 3, 6, 4, 5 members, respectively.

Let us select a family and note down the number of members in the family denoting X. Clearly, X is a random variable defined as below :

X(f1) = 3, X(f2) = 4, X(f3) = 3, X(f4) = 2, X(f5) = 5,

X(f6) = 4, X(f7) = 3, X(f8) = 6, X(f9) = 4, X(f10) = 5

Thus, X can take any value 2,3,4,5 or 6 depending upon which family is selected.

Now, X will take the value 2 when the family f4 is selected. X can take the value 3 when any one of the families f1, f3, f7 is selected.

Similarly, X = 4, when family f2, f6 or f9 is selected,

X = 5, when family f5 or f10 is selected

and X = 6, when family f8 is selected.

Since we had assumed that each family is equally likely to be selected, the probability that family f4 is selected is .

Thus, the probability that X can take the value 2 is . We write P(X = 2) =

Also, the probability that any one of the families f1, f3 or f7 is selected is

P({f1, f3, f7}) =

Thus, the probability that X can take the value 3 =

We write P(X = 3) =

Similarly, we obtain

P(X = 4) = P({f2, f6, f9}) =

P(X = 5) = P({f5, f10}) =

and P(X = 6) = P({f8}) =

Such a description giving the values of the random variable along with the corresponding probabilities is called the probability distribution of the random variable X.

In general, the probability distribution of a random variable X is defined as follows:

Definition 5 The probability distribution of a random variable X is the system of numbers

X : x1 x2 ... xn

P(X) : p1 p2 ... pn

where, = 1, i = 1, 2,..., n

The real numbers x1, x2,..., xn are the possible values of the random variable X and pi (i = 1,2,..., n) is the probability of the random variable X taking the value xi i.e.,
P(X = xi) = pi

 Note If xi is one of the possible values of a random variable X, the statement X = xi is true only at some point (s) of the sample space. Hence, the probability that X takes value xi is always nonzero, i.e. P(X = xi) ≠ 0.

Also for all possible values of the random variable X, all elements of the sample space are covered. Hence, the sum of all the probabilities in a probability distribution must be one.

Example 24 Two cards are drawn successively with replacement from a well-shuffled deck of 52 cards. Find the probability distribution of the number of aces.

Solution The number of aces is a random variable. Let it be denoted by X. Clearly, X can take the values 0, 1, or 2.

Now, since the draws are done with replacement, therefore, the two draws form independent experiments.

Therefore, P(X = 0) = P(non-ace and non-ace)

= P(non-ace) × P(non-ace)

=

P(X = 1) = P(ace and non-ace or non-ace and ace)

= P(ace and non-ace) + P(non-ace and ace)

= P(ace). P(non-ace) + P (non-ace) . P(ace)

=

and P(X = 2) = P (ace and ace)

=

Thus, the required probability distribution is

Example 25 Find the probability distribution of number of doublets in three throws of a pair of dice.

Solution Let X denote the number of doublets. Possible doublets are

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

Clearly, X can take the value 0, 1, 2, or 3.

Probability of getting a doublet

Probability of not getting a doublet

Now P(X = 0) = P (no doublet) =

P(X = 1) = P (one doublet and two non-doublets)

=

=

P(X = 2) = P (two doublets and one non-doublet)

=

and P(X = 3) = P (three doublets)

=

Thus, the required probability distribution is

Verification Sum of the probabilities

=

=

Example 26 Let X denote the number of hours you study during a randomly selected school day. The probability that X can take the values x, has the following form, where k is some unknown constant.

P(X = x) =

(a) Find the value of k.

(b) What is the probability that you study at least two hours ? Exactly two hours? At most two hours?

Solution The probability distribution of X is

(a) We know that = 1

Therefore 0.1 + k + 2k + 2k + k = 1

i.e. k = 0.15

(b) P(you study at least two hours) = P(X ≥ 2)

= P(X = 2) + P (X = 3) + P (X = 4)

= 2k + 2k + k = 5k = 5 × 0.15 = 0.75

P(you study exactly two hours) = P(X = 2)

= 2k = 2 × 0.15 = 0.3

P(you study at most two hours) = P(X ≤ 2)

= P (X = 0) + P(X = 1) + P(X = 2)

= 0.1 + k + 2k = 0.1 + 3k = 0.1 + 3 × 0.15 = 0.55

13.6.2 Mean of a random variable

In many problems, it is desirable to describe some feature of the random variable by means of a single number that can be computed from its probability distribution. Few such numbers are mean, median and mode. In this section, we shall discuss mean only. Mean is a measure of location or central tendency in the sense that it roughly locates a middle or average value of the random variable.

Definition 6 Let X be a random variable whose possible values x1, x2, x3, ..., xn occur with probabilities p1, p2, p3,..., pn, respectively. The mean of X, denoted by µ, is the number i.e. the mean of X is the weighted average of the possible values of X, each value being weighted by its probability with which it occurs.

The mean of a random variable X is also called the expectation of X, denoted by E(X).

In other words, the mean or expectation of a random variable X is the sum of the products of all possible values of X by their respective probabilities.

Example 27 Let a pair of dice be thrown and the random variable X be the sum of the numbers that appear on the two dice. Find the mean or expectation of X.

Solution The sample space of the experiment consists of 36 elementary events in the form of ordered pairs (xi, yi), where xi = 1, 2, 3, 4, 5, 6 and yi = 1, 2, 3, 4, 5, 6.

The random variable X i.e. the sum of the numbers on the two dice takes the values 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 or 12.

Now P(X = 2) = P({(1,1)})

P(X = 3) = P({(1,2), (2,1)})

P(X = 4) = P({(1,3), (2,2), (3,1)})

P(X = 5) = P({(1,4), (2,3), (3,2), (4,1)})

P(X = 6) = P({(1,5), (2,4), (3,3), (4,2), (5,1)})

P(X = 7) = P({(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)})

P(X = 8) = P({(2,6), (3,5), (4,4), (5,3), (6,2)})

P(X = 9) = P({(3,6), (4,5), (5,4), (6,3)})

P(X = 10) = P({(4,6), (5,5), (6,4)})

P(X = 11) = P({(5,6), (6,5)})

P(X = 12) = P({(6,6)})

The probability distribution of X is

Therefore,

µ = E(X) =

= = 7

Thus, the mean of the sum of the numbers that appear on throwing two fair dice is 7.

13.6.3 Variance of a random variable

The mean of a random variable does not give us information about the variability in the values of the random variable. In fact, if the variance is small, then the values of the random variable are close to the mean. Also random variables with different probability distributions can have equal means, as shown in the following distributions of X and Y.

Clearly E(X) =

and E(Y) =

The variables X and Y are different, however their means are same. It is also easily observable from the diagramatic representation of these distributions (Fig 13.5).

Fig 13.5

To distinguish X from Y, we require a measure of the extent to which the values of the random variables spread out. In Statistics, we have studied that the variance is a measure of the spread or scatter in data. Likewise, the variability or spread in the values of a random variable may be measured by variance.

Definition 7 Let X be a random variable whose possible values x1, x2,...,xn occur with probabilities p(x1), p(x2),..., p(xn) respectively.

Let µ = E (X) be the mean of X. The variance of X, denoted by Var (X) or is defined as

=

or equivalently = E(X – µ)2

The non-negative number

σx =

is called the standard deviation of the random variable X.

Another formula to find the variance of a random variable. We know that,

or Var (X) =

or Var (X) = E(X2) – [E(X)]2, where E(X2) =

Example 28 Find the variance of the number obtained on a throw of an unbiased die.

Solution The sample space of the experiment is S = {1, 2, 3, 4, 5, 6}.

Let X denote the number obtained on the throw. Then X is a random variable which can take values 1, 2, 3, 4, 5, or 6.

Also P(1) = P(2) = P(3) = P(4) = P(5) = P(6) =

Therefore, the Probability distribution of X is

=

Also E(X2) =

Thus, Var (X) = E (X2) – (E(X))2

=

Example 29 Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean, variance and standard deviation of the number of kings.

Solution Let X denote the number of kings in a draw of two cards. X is a random variable which can assume the values 0, 1 or 2.

Now P(X = 0) = P (no king)

P(X = 1) = P (one king and one non-king)

=

and P(X = 2) = P (two kings) =

Thus, the probability distribution of X is

=

Also E(X2) =

=

Now Var(X) = E(X2) – [E(X)]2

=

Therefore σx =

13.7 Bernoulli Trials and Binomial Distribution

13.7.1 Bernoulli trials

Many experiments are dichotomous in nature. For example, a tossed coin shows a ‘head’ or ‘tail’, a manufactured item can be ‘defective’ or ‘non-defective’, the response to a question might be ‘yes’ or ‘no’, an egg has ‘hatched’ or ‘not hatched’, the decision is ‘yes’ or ‘no’ etc. In such cases, it is customary to call one of the outcomes a ‘success’ and the other ‘not success’ or ‘failure’. For example, in tossing a coin, if the occurrence of the head is considered a success, then occurrence of tail is a failure.

Each time we toss a coin or roll a die or perform any other experiment, we call it a trial. If a coin is tossed, say, 4 times, the number of trials is 4, each having exactly two outcomes, namely, success or failure. The outcome of any trial is independent of the outcome of any other trial. In each of such trials, the probability of success or failure remains constant. Such independent trials which have only two outcomes usually referred as ‘success’ or ‘failure’ are called Bernoulli trials.

Definition 8 Trials of a random experiment are called Bernoulli trials, if they satisfy the following conditions :

(i) There should be a finite number of trials.

(ii) The trials should be independent.

(iii) Each trial has exactly two outcomes : success or failure.

(iv) The probability of success remains the same in each trial.

For example, throwing a die 50 times is a case of 50 Bernoulli trials, in which each trial results in success (say an even number) or failure (an odd number) and the probability of success (p) is same for all 50 throws. Obviously, the successive throws of the die are independent experiments. If the die is fair and have six numbers 1 to 6 written on six faces, then p = and q = 1 – p = = probability of failure.

Example 30 Six balls are drawn successively from an urn containing 7 red and 9 black balls. Tell whether or not the trials of drawing balls are Bernoulli trials when after each draw the ball drawn is

(i) replaced (ii) not replaced in the urn.

Solution

(i) The number of trials is finite. When the drawing is done with replacement, the probability of success (say, red ball) is p = which is same for all six trials (draws). Hence, the drawing of balls with replacements are Bernoulli trials.

(ii) When the drawing is done without replacement, the probability of success
(i.e., red ball) in first trial is , in 2nd trial is if the first ball drawn is red or if the first ball drawn is black and so on. Clearly, the probability of success is not same for all trials, hence the trials are not Bernoulli trials.

13.7.2 Binomial distribution

Consider the experiment of tossing a coin in which each trial results in success (say, heads) or failure (tails). Let S and F denote respectively success and failure in each trial. Suppose we are interested in finding the ways in which we have one success in six trials.

Clearly, six different cases are there as listed below:

SFFFFF, FSFFFF, FFSFFF, FFFSFF, FFFFSF, FFFFFS.

Similarly, two successes and four failures can have combinations. It will be lengthy job to list all of these ways. Therefore, calculation of probabilities of 0, 1, 2,..., n number of successes may be lengthy and time consuming. To avoid the lengthy calculations and listing of all the possible cases, for the probabilities of number of successes in n-Bernoulli trials, a formula is derived. For this purpose, let us take the experiment made up of three Bernoulli trials with probabilities p and q = 1 – p for success and failure respectively in each trial. The sample space of the experiment is the set

S = {SSS, SSF, SFS, FSS, SFF, FSF, FFS, FFF}

The number of successes is a random variable X and can take values 0, 1, 2, or 3. The probability distribution of the number of successes is as below :

P(X = 0) = P(no success)

= P({FFF}) = P(F) P(F) P(F)

= q . q . q = q3 since the trials are independent

P(X = 1) = P(one successes)

= P({SFF, FSF, FFS})

= P({SFF}) + P({FSF}) + P({FFS})

= P(S) P(F) P(F) + P(F) P(S) P(F) + P(F) P(F) P(S)

= p.q.q + q.p.q + q.q.p = 3pq2

P(X = 2) = P (two successes)

= P({SSF, SFS, FSS})

= P({SSF}) + P ({SFS}) + P({FSS})

= P(S) P(S) P(F) + P(S) P(F) P(S) + P(F) P(S) P(S)

= p.p.q. + p.q.p + q.p.p = 3p2q

and P(X = 3) = P(three success) = P ({SSS})

= P(S) . P(S) . P(S) = p3

Thus, the probability distribution of X is

Also, the binominal expansion of (q + p)3 is

q3 + 3q2p + 3qp2 + p3

Note that the probabilities of 0, 1, 2 or 3 successes are respectively the 1st, 2nd, 3rd and 4th term in the expansion of (q + p)3.

Also, since q + p = 1, it follows that the sum of these probabilities, as expected, is 1.

Thus, we may conclude that in an experiment of n-Bernoulli trials, the probabilities of 0, 1, 2,..., n successes can be obtained as 1st, 2nd,...,(n + 1)th terms in the expansion of (q + p)n. To prove this assertion (result), let us find the probability of x-successes in an experiment of n-Bernoulli trials.

Clearly, in case of x successes (S), there will be (n – x) failures (F).

Now, x successes (S) and (n – x) failures (F) can be obtained in ways.

In each of these ways, the probability of x successes and (n − x) failures is

= P(x successes) . P(n–x) failures is

Thus, the probability of x successes in n-Bernoulli trials is px qn–x
or nCx px qn–x

Thus P(x successes) = , x = 0, 1, 2,...,n. (q = 1 – p)

Clearly, P(x successes), i.e. is the (x + 1)th term in the binomial expansion of (q + p)n.

Thus, the probability distribution of number of successes in an experiment consisting of n Bernoulli trials may be obtained by the binomial expansion of (q + p)n. Hence, this distribution of number of successes X can be written as

The above probability distribution is known as binomial distribution with parameters n and p, because for given values of n and p, we can find the complete probability distribution.

The probability of x successes P(X = x) is also denoted by P(x) and is given by P(x) = nCx qn–xpx, x = 0, 1,..., n. (q = 1 – p)

This P(x) is called the probability function of the binomial distribution.

A binomial distribution with n-Bernoulli trials and probability of success in each trial as p, is denoted by B(n, p).

Let us now take up some examples.

Example 31 If a fair coin is tossed 10 times, find the probability of

(i) exactly six heads

(ii) at least six heads

(iii) at most six heads

Solution The repeated tosses of a coin are Bernoulli trials. Let X denote the number of heads in an experiment of 10 trials.

Clearly, X has the binomial distribution with n = 10 and p =

Therefore P(X = x) = nCxqn–xpx, x = 0, 1, 2,...,n

Here n = 10, , q = 1 – p =

Therefore P(X = x) =

Now (i) P(X = 6) =

(ii) P(at least six heads) = P(X ≥ 6)

= P (X = 6) + P (X = 7) + P (X = 8) + P(X = 9) + P (X = 10)

=

=

(iii) P(at most six heads) = P(X ≤ 6)

= P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3)

+ P (X = 4) + P (X = 5) + P (X = 6)

=

+

=

Example 32 Ten eggs are drawn successively with replacement from a lot containing 10% defective eggs. Find the probability that there is at least one defective egg.

Solution Let X denote the number of defective eggs in the 10 eggs drawn. Since the drawing is done with replacement, the trials are Bernoulli trials. Clearly, X has the binomial distribution with n = 10 and .

Therefore q =

Now P(at least one defective egg) = P(X ≥ 1) = 1 – P (X = 0)

= =

Miscellaneous Examples

Example 33 Coloured balls are distributed in four boxes as shown in the following table:

A box is selected at random and then a ball is randomly drawn from the selected box. The colour of the ball is black, what is the probability that ball drawn is from the box III?

Solution Let A, E1, E2, E3 and E4 be the events as defined below :

A : a black ball is selected E1 : box I is selected

E2 : box II is selected E3 : box III is selected

E4 : box IV is selected

Since the boxes are chosen at random,

Therefore P(E1) = P(E2) = P(E3) = P(E4) =

Also P(A|E1) = , P(A|E2) = , P(A|E3) = and P(A|E4) =

P(box III is selected, given that the drawn ball is black) = P(E3|A). By Bayes' theorem,

P(E3|A) =

=

Example 34 Find the mean of the Binomial distribution .

Solution Let X be the random variable whose probability distribution is 

Here n = 4, p = and q =

We know that P(X = x) = , x = 0, 1, 2, 3, 4.

i.e. the distribution of X is

Now Mean (µ) =

= +

=

=

Example 35 The probability of a shooter hitting a target is . How many minimum number of times must he/she fire so that the probability of hitting the target at least once is more than 0.99?

Solution Let the shooter fire n times. Obviously, n fires are n Bernoulli trials. In each trial, p = probability of hitting the target = and q = probability of not hitting the target = . Then P(X = x) = .

Now, given that,

P(hitting the target at least once) > 0.99

i.e. P(x ≥ 1) > 0.99

Therefore, 1 – P (x = 0) > 0.99

or > 0.99

or < 0.01

or 4n > = 100 ... (1)

The minimum value of n to satisfy the inequality (1) is 4.

Thus, the shooter must fire 4 times.

Example 36 A and B throw a die alternatively till one of them gets a ‘6’ and wins the game. Find their respective probabilities of winning, if A starts first.

Solution Let S denote the success (getting a ‘6’) and F denote the failure (not getting a ‘6’).

Thus, P(S) =

P(A wins in the first throw) = P(S) =

A gets the third throw, when the first throw by A and second throw by B result into failures.

Therefore, P(A wins in the 3rd throw) = P(FFS) =

=

P(A wins in the 5th throw) = P (FFFFS) and so on.

Hence, P(A wins) = 

= =

P(B wins) = 1 – P (A wins) =

Remark If a + ar + ar2 + ... + arn–1 + ..., where |r| < 1, then sum of this infinite G.P. is given by (Refer A.1.3 of Class XI Text book).

Example 37 If a machine is correctly set up, it produces 90% acceptable items. If it is incorrectly set up, it produces only 40% acceptable items. Past experience shows that 80% of the set ups are correctly done. If after a certain set up, the machine produces 2 acceptable items, find the probability that the machine is correctly setup.

Solution Let A be the event that the machine produces 2 acceptable items.

Also let B1 represent the event of correct set up and B2 represent the event of incorrect setup.

Now P(B1) = 0.8, P(B2) = 0.2

P(A|B1) = 0.9 × 0.9 and P(A|B2) = 0.4 × 0.4

Therefore P(B1|A) =

=