There are 2 bags each containing 3 white and 5 black balls and 4 bags each containing 6 white and 4 black…
There are 2 bags each containing 3 white and 5 black balls and 4 bags each containing 6 white and 4 black balls. If a ball drawn randomly from a bag is found to be black, then the probability that this ball is from the first set of bags is
$\frac{25}{57}$
$\frac{25}{41}$
$\frac{2}{5}$
$\frac{3}{5}$
Solution
Let, $B_1=$ Bags of first group, $B_2=$ Bags of second group
$P\left(B_1\right)=\frac{2}{6}=\frac{1}{3}, P\left(B_2\right)=\frac{4}{6}=\frac{2}{3}$
and let $B=$ the event that the ball is black
$P\left(\frac{B}{B_1}\right)=\frac{5}{8}, P\left(\frac{B}{B_2}\right)=\frac{2}{5}$
Now, $P\left(\frac{B_1}{B}\right)=\frac{P\left(B_1\right) P\left(\frac{B}{B_1}\right)}{P\left(B_1\right) P\left(\frac{B}{B_1}\right)+P\left(B_2\right) P\left(\frac{B}{B_2}\right)}$
$=\frac{\frac{1}{3} \times \frac{5}{8}}{\frac{1}{3} \times \frac{5}{8}+\frac{2}{3} \times \frac{2}{5}}=\frac{\frac{5}{8}}{\frac{5}{8}+\frac{4}{5}}=\frac{25}{57}$