Conditional Probability

The probability of occurrence of an event B when it is known that some event A has occurred is called a condition probability and is denoted by P(B/A). The symbol P(B/A) is usually read '‘the probability that B occurs given that A occurs” or ”simply probability of B, given A”.

Consider two events ‘A’ and ‘B’ of sample-space S. When it is known that event ‘A’ has occurred, it means that sample space would reduce to the sample points representing event A. Now for P(B/A) we must look for the sample points representing the simultaneous occurrence of A and B i.e. sample points in AÇB.

&⇒ P(B/A) =

Thus P(B/A) = , where 0 < P(A) £ 1

Similarly, P(A/B) = 0 < P(B) £ 1

Hence, P(A Ç B) =

Consider the event ‘B’ of getting a ‘4’ when a fair die is tossed. Now suppose that it is known that toss of die resulted in a number greater than 3 (say event A). And we have to obtain P(B/A) i.e. the probability of getting a ‘4’ given that a number greater than 3
has occurred. .

Clearly A = {4, 5, 6}, B = {4}
&⇒ P(B/A) =

also P(AÇB) = and P(A) =

&⇒ P(B/A) = =

Example.1

The probability that a married man watches a certain T.V. show is 0.4 and the probability that a married woman watches the show is 0.5. The probability that a man watches the show, given that his wife does, is 0.7. Find .

(a) the probability that married couple watch the show

(b) the probability that a wife watches the show given that her husband does.

Solution:

Let ‘H’ be the event that married man watches the show and ‘W’ be the event that married woman watch the show,

&⇒ P(H) = 0.4, P(W) = 0.5, P(H/W) = 0.7.

(a) P(HÇW) = P(W)×P(H/W) = 0.5 ´ 0.7 = 0.35.

(b) P(W/H) =

= 0.4 + 0.5 - 0.35 = 0.55.

Example.2

Consider the sample space ‘S’ representing the adults in a small town who have completed the requirements for a college degree. They have been categorised according to sex and employment as under:.

	Employed	Unemployed
Male	460	40
Female	140	260

A employed person is selected at random. Find the probability that chosen one is male.

Solution

Let M be the event that man is chosen and E be the event that chosen one is employed.

From the concept of reduced sample space we immediately get

P(M/E) =

This can be done also as P(E) =

P(EÇM) =

&⇒ P(M/E) =

Example.3

A bag contains 3 white balls and 2 black balls, another contains 5 white and 3 black balls. If a bag is chosen at random and a ball is drawn from it. What is the probability that it is white.

Solution:

The probability that the first bag is chosen is 1/2 and the chance of drawing a white ball from it is 3/5.

\ Chance of choosing the first bag and drawing a white ball is 1/2.3/5.

Similarly, the chance of choosing the second bag and drawing a white ball is 1/2.5/8.

\ The chance of randomly choosing a bag and drawing a white ball is = (Mutually exclusive cases) = 49/80.

Independent Events:

Two events A and B are said to be independent if occurrence or non-occurrence of one does not affect the occurrence or non-occurrence of the other,

i.e. P(B/A) = P(B), P(A) > 0 similarly P(A/B) = P(A), P(B) > 0.

&⇒ P(B/A) =

&⇒ P(AÇB) = P(A)×P(B)

If the events are not independent, they are said to be dependent.

Remarks:

q If P(A) = 0
&⇒ for event ‘B’, 0 £ P(AÇB) £ P(A).

&⇒ P(AÇB) = 0, thus P(AÇB) = P(A)×P(B) = 0. Hence an impossible event would be independent of any other event.

q Distinction between independent and mutually exclusive events must be carefully made. If A and B are two mutually exclusive and possible events of sample space ‘S’ then P(A) > 0, P(B) > 0 and P(AÇB) = 0 > P(A)×P(B) so that A and B can’t be independent. Infact P(A/B) = 0 similarly P(B/A) = 0. Consequently, mutually exclusive events are strongly dependent. .

q Two events A and B are independent if and only if A and B¢ are independent or A¢ and B are independent or A¢ and B¢ are independent. .

We have P(AÇB) = P(A)×P(B)

Now P(AÇB¢) = P(A) - P(AÇB)

= P(A) - P(A)×P(B)

= P(A) (1-P(B))

= P(A)×P(B¢)

Thus A and B¢ are independent.

Similarly P(A¢ÇB) = P(B) - P(AÇB)

= P(B)×P(A¢)

Finally P(A¢ Ç B¢) = 1 - P(AÈB)

= 1 - P(A) - P(B) + P(AÇB)

= 1 - P(A) - P(B) + P(A)×P(B)

= (1-P(A)) - P(B) (1-P(A))

= (1-P(A)) (1-P(B))

= P(A¢)×P(B¢)

Thus A¢ and B¢ are also independent.

Example.1

An event A₁ can happen with probability p₁ and event A₂ can happen with probability p₂.

What is the probability that

(i) exactly one of them happens

(ii) at least one of them happens(Given A₁ and A₂ are independent events).

Solution:

(i) The probability that A₁ happens is p₁

\ The probability that A₁ fails is 1 - p₁

Also the probability that A₂ happens is p₂

Now, the chance that A₁ happens and A₂ fails is p₁(1 - p₂) and the chance that A₁ fails and A₂ happens is p₂(1-p₁)

\ The probability that one and only one of them happens is

p₁(1 - p₂) + p₂(1 - p₁) = p₁ + p₂ - 2p₁p₂

(ii) The probability that both of them fail = (1 - p₁) (1 - p₂).

\ probability that atleast one happens=1 -(1 - p₁) (1 - p₂)=p₁+p₂ - p₁ p₂

Mutual Independence and Pairwise Independence:

Three events A, B, C are said to be mutually independent if,

P(AÇB) = P(A)×P(B), P(A Ç C) = P(A)×P(C), P(B Ç C) = P(B)×P(C) and P(AÇBÇC) = P(A)×P(B)×P(C)

These events would be said to be pairwise independent if,

P(AÇB) = P(A)×P(B), P(BÇC) = P(B)×P(C) and P(AÇC) = P(A)×P(C).

Thus mutually independent events are pairwise independent but the converse may not be true.

Example.1

A lot contains 50 defective and 50 non-defective bulbs. Two bulbs are drawn at random, one at a time with replacement. The events A, B, C are defined as under;.

A = {the first bulb is defective}

B = {the second bulb is non defective}

C = {the two bulbs are both defective or both are non-defective}.

Categorise the events A, B, C to be pairwise independent or mutually independent. .

Solution

Let D_i (N_i) denote the occurrence of a defective (non-defective) bulb at the i^th draw (i = 1, 2). Now sample-space ‘S’ is; .

S = {D₁ D₂, D₁ N₂, N₁ D₂, N₁ N₂}
&⇒ P(D_i) = (i = 1, 2)

A = {D₁ D₂, D₁ N₂}, B = {D₁ N₂, N₁ N₂} , C = {D₁ D₂, N₁ N₂}

&⇒ AÇB = {D₁ N₂}, AÇC = {D₁ D₂}, BÇC = {N₁ N₂} and A Ç B Ç C = f

P(A) = P(D₁ D₂) + P(D₁N₂) =

P(B) = P(D₁ N₂) + P(N₁ N₂) =

P(C) = P(D₁ D₂) + P(N₁N₂) =

P(AÇB) = P(D₁ N₂) == P(A)×P(B)

P(BÇC) = P(N₁ N₂) == P(B)×P(C)

P(AÇC) = P(D₁ D₂) = = P(A)×P(C)

And P(A Ç B Ç C) = 0 > P(A) × P(B) × P(C)

&⇒ Events are not mutually independent but are pairwise independent.

Example.2

A person draws a card from a pack of 52, replaces it and shuffles it. He continues doing it until he draws a spade. What is the chance that he has to make .

(i) atleast 3 trials

(ii) exactly 3 trials :

Solution

(i) For atleast 3 trials, he has to fail at the first 2 attempts and then after that it doesn't make a difference if he fails or wins at the 3rd or the subsequent attempts.

Chance of success at any attempt = 1/4

\ chance of failure = 3/4

\ chance of failing in first 2 attempt =

(ii) For exactly 3 attempts, he has to fail in the first two attempts and succeed in the 3rd attempt.

\ required probability =

Basic Concepts Of Probability Part 4

Conditional Probability

Example.1

Solution:

Example.2

Solution

Example.3

Solution:

Independent Events:

Remarks:

Example.1

Solution:

Mutual Independence and Pairwise Independence:

Example.1

Solution

Example.2

Solution

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