# Conditional Probability

Let A and B be the two events associated with a random experiment. Then, the probability of occurrence of A under the condition that B has already occurred and P (B) â‰ 0, is called the conditional probability and it is denoted by P(A/B).Thus, P(A/B) = Probability of occurrence of A given that B has already happened.
Similarly, P (B/A) = Probability of occurrence of B given that A has already happened.
Sometimes, P (A/B) is also used to denote the probability occurrence of A when B occurs.

Example-1

A bag contains 5 white and 4 red balls. Two balls are drawn form the bag one after the other without replacement. Consider the following events.
A = drawing a white ball in the first draw, B = drawing a red ball in the second draw.

Solution

Now, P(B/A) = Probability of drawing a red ball in the second draw given that a white ball has already been drawn in the first draw.

Since 8 balls are left after drawing a white ball in the first draw and out of these 8 balls 4 balls are red, therefore

P (B/A) =

**Note**that P (A/B) is not meaningful in this experiment because A cannot occur after the occurrence of B.

Example-2

One card is drawn from a pack of 52 cards, each of the 52 cards being equally likely to be drawn. Find the probability that the card drawn is

- a king
- either a red or a king
- a red and a king

Solution

Out of 52 cards, one card can be drawn in

^{52}C_{1}ways. Therefore, exhaustive number of cases =^{52}C_{1}= 52

- There are 4 kings in a pack of cards, out of which one can be drawn in
^{4}C_{1 }ways.^{4}C_{1}= 4, so the required probability = - There are 28 cards in a pack of cards which are either a red or a king. Therefore, one can be drawn in
^{28}C_{1}ways. Therefore, the favourable number of cases =^{28}C_{1}=28. - There are 2 cards which are red and king, i.e., red kings. Therefore, the favourable number of cases
^{2}C_{1}= 2. so the required probability =

Example-3

An urn contains 9 blue, 7 white and 4 black balls. If 2 balls are drawn at random, find the probability that

- Both the balls are blue
- One ball is white

Solution

There are 20 balls in the bag out of which 2 balls can be drawn in

^{20}C_{2}ways. So the total number of cases(sample space) =^{20}C_{2}= 190.- There are 9 blue balls out of which 2 balls can be drawn in
^{9}C_{2}ways. Therefore, the favourable number of cases =^{9}C_{2}= 36. So the required probability = - There are 7 white balls out of which one white can be drawn in
^{7}C_{1}ways. One ball from the remaining 13 balls can be drawn in^{13}C_{1}ways. Therefore, one white and one other colour ball can be drawn in^{7}C_{1}Ã—^{13}C_{1}ways. So the favourable number of cases =^{7}C_{1}Ã—^{13}C_{1}= 91.

Example-4

Three persons A, B and C are to speak at a function along with five others. If they all speak in random order, the probability that A speaks before B and B speaks before C is

- 3/8
- 1/6
- 3/5
- None of these

Solution

The total number of ways in which 8 persons can speak is

^{8}P_{8 }= 8!. The number of ways in which A, B and C can be arranged in the specified speaking order is^{8}C_{3}. There are 5! ways in which the other five can speak. So, the favourable number of ways is^{8}C_{3}= 5!.Hence, the required probability =

Example-5

There is a point inside a circle. What is the probability that this point is closer to the circumference than to the centre?

Solution

Assume that the radius of the bigger circle is r, and the radius of the inner circle is r/2. Point will be closer to the circumference than to the centre if the point is lying in the segment B.

Area of segment B = Â¾ Ï€r

^{2}So, the probability of point being closer to the circumference = Â¾ Ï€r

^{2}/Ï€r^{2 }= 3/4