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Question

Three boxes contain 6 red, 4 black; 5 red, 5 black and 4 red, 6 black balls respectively. One of the box is selected at random and a ball is drawn from it. If the ball drawn is red, then the probability that it is drawn from the first bag is k. The value of 10k is___.

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Solution

Let A, B, C and D be the events defined as follows:
A = first box is chosen
B = second box is chosen
C = third box is chosen
D = ball drawn is red.
Since there are three boxes and one of the three boxes is chosen at random, therefore
P(A)=P(B)=P(C)=13
If A has already occurred, then first box has been chosen which contains 6 red and 4 black balls. The probability of drawing a red ball from it is 610
So, P(DA)=610
Similary, P(DB)=510 and P(DC)=410
We are required to find P(AD)i.e, given that the drawn is red, what is the probability that it is drawn from the first box. By Baye's rule.
P(AD)=P(A).P(DA)P(A).P(DA)+P(B).P(DB)+P(C).P(DC)=13×61(13×610)+(13×510)+13×410=25

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