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___.

Answer 1

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
$P\left(A\right)=P\left(B\right)=P\left(C\right)=\frac{1}{3}$
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 $\frac{6}{10}$
So, $P\left(\frac{D}{A}\right)=\frac{6}{10}$
Similary, $P\left(\frac{D}{B}\right)=\frac{5}{10}andP\left(\frac{D}{C}\right)=\frac{4}{10}$
We are required to find $P\left(\frac{A}{D}\right)$i.e, given that the drawn is red, what is the probability that it is drawn from the first box. By Baye's rule.
$P\left(\frac{A}{D}\right)=\frac{P\left(A\right).P\left(\frac{D}{A}\right)}{P\left(A\right).P\left(\frac{D}{A}\right)+P\left(B\right).P\left(\frac{D}{B}\right)+P\left(C\right).P\left(\frac{D}{C}\right)}=\frac{\frac{1}{3}\times \frac{6}{1}}{(\frac{1}{3}\times \frac{6}{10})+(\frac{1}{3}\times \frac{5}{10})+\frac{1}{3}\times \frac{4}{10}}=\frac{2}{5}$