Question

A bag contains $4$ red and $6$ black balls and another bag contains $3$ red and $5$ black balls. One of the two bags is selected at random and a ball is drawn from the bag which is found to be red. The probability that the ball is drawn from the second bag is $\frac{15}{P}$. Find the value of $P$.

Answer 1

Given first bag has $4$ red and $6$ black balls

second bag has $3$ red and $5$ black balls

The probability of selecting first bag is equal to the probability of selecting second bag $=\frac{1}{2}$

The probability of selecting red ball from first bag is $\frac{4}{6+4}=\frac{4}{10}=\frac{2}{5}$

The probability of selecting red ball from second bag is $\frac{3}{3+5}=\frac{3}{8}$

From bayes law , the probability that the red ball is drawn from second bag is $\frac{\frac{1}{2}\times \frac{3}{8}}{\frac{1}{2}\times \frac{3}{8}+\frac{1}{2}\times \frac{2}{5}}=\frac{\frac{3}{8}}{\frac{3}{8}+\frac{2}{5}}=\frac{15}{31}$

the probability that the red ball is drawn from second bag is$\frac{15}{31}=\frac{15}{P}$

by comparing we can say that the value of $P=31$