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Basic Concepts Of Probability Part 4

Posted on - 04-01-2017

Math

IIT JEE

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.

(c) the probability that atleast one person of a married couple will watch the show.

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) =

(c) P(HÈW) = P(H) + P(W) - P(HÇW)

= 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 Di (Ni) denote the occurrence of a defective (non-defective) bulb at the ith draw (i = 1, 2). Now sample-space ‘S’ is; .

S = {D1 D2, D1 N2, N1 D2, N1 N2}
&⇒
P(Di) = (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 =

 
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