Question

Bag I contains $3$ red and $4$ black balls and Bag II contains $4$ red and $5$ black balls. Two balls are transferred at random from Bag I to Bag II and then a ball is drawn from Bag II. The ball so drawn is found to be red in colour. Find the probability that the transferred balls were both black

Answer 1

Let $E_1,E_2,E_3$ and $A$ be events such that

$E_1$ = Both transferred ball from Bag I to bag II are red.

$E_2$ = Both transferred ball from Bag I to bag II are black.

$E_3$ = Out of two transferred ball one is red and other is black.

$A$ = drawing a red ball from Bag II.

Here, $P\left(\frac{E_2}{A}\right)$ is required.

Now, $P\left(E_1\right)=\frac{3C_2}{7C_2}=\frac{3!}{2!1!}\times \frac{2!\times 5!}{7!}=\frac{1}{7}$

$P\left(E_2\right)=\frac{4C_2}{7C_2}=\frac{4!}{2!2!}\times \frac{2!\times 5!}{7!}=\frac{2}{7}$

$P\left(E_3\right)=\frac{3C_1\times 4C_1}{7C_2}=\frac{3!\times 4!}{7!}\times \frac{2!\times 5!}{1!}=\frac{4}{7}$

$P\left(\frac{A}{E_1}\right)=\frac{6}{11},P\left(\frac{A}{E_2}\right)=\frac{4}{11},P\left(\frac{A}{E_3}\right)=\frac{5}{11}$

Therefore, $P\left(\frac{E_2}{A}\right)=\frac{P\left(E_2\right).P\left(\frac{A}{E_2}\right)}{P\left(E_1\right).\left(\frac{A}{E_1}\right)+P\left(E_2\right).\left(\frac{A}{E_2}\right)+P\left(E_3\right).\left(\frac{A}{E_3}\right)}$

$aftersolving$

$=\frac{\frac{8}{77}}{\frac{6}{77}+\frac{8}{77}+\frac{20}{77}}=\frac{8}{77}\times \frac{77}{34}=\frac{4}{17}$

Therefore, the probability that the transferred balls were both black = $\frac{4}{17}$