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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) ≠ 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.
 
Similarly, P(B/A) is used to denote the probability of occurrence of B when A occurs.
 
Following examples illustrate the various meanings of these notations:
 
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) = Description: 1635.png 
 
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
  1. a king
  2. either a red or a king
  3. a red and a king
Solution
Out of 52 cards, one card can be drawn in 52C1 ways. Therefore, exhaustive number of cases = 52C1 
= 52
  1. There are 4 kings in a pack of cards, out of which one can be drawn in 4C1 ways.
     
    Therefore, the favourable number of cases = 4C1 = 4, so the required probability = Description: 1644.png
  2. There are 28 cards in a pack of cards which are either a red or a king. Therefore, one can be drawn in 28C1 ways. Therefore, the favourable number of cases = 28C1=28.
     
    So the required probability = Description: 1653.png
  3. There are 2 cards which are red and king, i.e., red kings. Therefore, the favourable number of cases
     
    2C1 = 2. so the required probability = Description: 1662.png
 
Example-3
An urn contains 9 blue, 7 white and 4 black balls. If 2 balls are drawn at random, find the probability that
  1. Both the balls are blue
  2. One ball is white
Solution
There are 20 balls in the bag out of which 2 balls can be drawn in 20C2 ways. So the total number of cases(sample space) = 20C2 = 190.
  1. There are 9 blue balls out of which 2 balls can be drawn in 9C2 ways. Therefore, the favourable number of cases = 9C2 = 36. So the required probability = Description: 1671.png
  2. There are 7 white balls out of which one white can be drawn in 7C1 ways. One ball from the remaining 13 balls can be drawn in 13C1 ways. Therefore, one white and one other colour ball can be drawn in 7C1 × 13C1 ways. So the favourable number of cases = 7C1 × 13C1 = 91.
     
    So, the required probability = Description: 1681.png
 
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
  1. 3/8
  2. 1/6
  3. 3/5
  4. None of these
Solution
The total number of ways in which 8 persons can speak is 8P8 = 8!. The number of ways in which A, B and C can be arranged in the specified speaking order is 8C3. There are 5! ways in which the other five can speak. So, the favourable number of ways is 8C3 = 5!.
 
Hence, the required probability = Description: 1690.png 
 
 
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 = ¾ πr2
 
So, the probability of point being closer to the circumference = ¾ πr2/πr2 = 3/4
 
Description: 14-1.tif
 




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