Conditional Probability

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 A$\cap$B.

P(B/A) = $\frac{n(A\cap B)}{n(A)}=\frac{\frac{n(A\cap B)}{n(S)}}{\frac{n(A)}{n(S)}}=\frac{P(A\cap B)}{P(A)}$

Thus P(B/A) = $\frac{P(A\cap B)}{P(A)}$, where 0 < P(A) $⩽$1

Similarly, P(A/B) = $\frac{P(A\cap B)}{P(B)},$0 < P(B) $⩽$1

Hence, P(A $\cap$ B) = $\{\begin{array}{c}P(A)\cdot P(B/A),P(A)>0\\ P(B)\cdot P(A/B),P(B)>0\end{array}$

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} $\Rightarrow$ P(B/A) = $\frac{1}{3}$

also P(A$\cap$B) = $\frac{1}{6}$ and P(A) = $\frac{3}{6}=\frac{1}{2}$ $\Rightarrow$ P(B/A) = $\frac{P(A\cap B)}{P(A)}$ = $\frac{1/6}{1/2}=\frac{1}{3}$

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.

Sol: Let 'H' be the event that married man watches the show and 'W' be the probability that married woman watch the show,

$\Rightarrow$ P(H) = 0.4, P(W) = 0.5, P(H/W) = 0.7

(a) P(H$\cap$W) = P(W)×P(H/W) = 0.5 $\times$ 0.7 = 0.35

(b) P(W/H) = $\frac{P(H\cap W)}{P(H)}=\frac{0.35}{0.4}=\frac{7}{8}$

(c) P(H$\cup$W) = P(H) + P(W) - P(H$\cap$W)

= 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.

Sol: 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) = $\frac{460}{600}=\frac{23}{30}$

Also, P(E) = $\frac{600}{900}={}^2/{}_3$

P(E$\cap$M) = $\frac{460}{900}=\frac{23}{45}$ $\Rightarrow$ P(M/E) = $\frac{23/45}{2/3}=\frac{23}{30}$

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) $\ne$]0 similarly P(A/B) = P(A), P(B) $\ne$]0

$\Rightarrow$ P(B/A) = $\frac{P(A\cap B)}{P(A)}=P(B)$ $\Rightarrow$ P(A$\cap$B) = P(A)×P(B)

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

Remarks:

$\bullet$ If P(A) = 0 $\Rightarrow$ for event 'B', 0 $⩽$P(A$\cap$B) $⩽$P(A).

$\Rightarrow$ P(A$\cap$B) = 0, thus P(A$\cap$B) = P(A)×P(B) = 0. Hence an impossible event would be independent of any other event.

$\bullet$ Distinction between independent and mutually exclusive events must be carefully made. Independence is a property of probability whereas the mutual exclusion is a set-theoretic concept. If A and B are two mutually exclusive and possible events of sample space 'S' then P(A) > 0, P(B) > 0 and P(A$\cap$B) = 0 $\ne$]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.

$\bullet$ 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(A$\cap$B) = P(A)×P(B)

Now P(A$\cap$B') = P(A) - P(A$\cap$B) = P(A) - P(A)×P(B) = P(A) (1-P(B)) = P(A)×P(B')

Thus A and B' are independent.

Similarly P(A'$\cap$B) = P(B) - P(A$\cap$B) = P(B)×P(A')

Finally P(A' $\cap$ B') = 1 - P(A$\cup$B) = 1 - P(A) - P(B) + P(A$\cap$B)

= 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 -3: 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).

Sol: (i) The probability that A₁ happens is p₁

$\therefore$ 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)

$\therefore$ 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).

$\therefore$ 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(A$\cap$B) = P(A)×P(B), P(A $\cap$ C) = P(A)×P(C), P(B $\cap$ C) = P(B)×P(C)

and P(A$\cap$B$\cap$C) = P(A)×P(B)×P(C)

These events would be said to be pairwise independent if,

P(A$\cap$B) = P(A)×P(B), P(B$\cap$C) = P(B)×P(C) and P(A$\cap$C) = P(A)×P(C).

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