Question

There are two bags, one of which contains $3$ black and $4$ white balls,$II$ bag contain $4$ black and $3$ white balls.A die is cast, if the number $3$ or less than $3$ turns up, a ball is drawn from $1$ bag and if the face greater than $3$ turns up, a ball is drawn from $II$ bag. It is found that ball drawn is black. Find the probability that it is from bag I.

• A. $\frac{3}{14}$
• B. $\frac{4}{7}$
• C. $\frac{2}{7}$
• D. $\frac{3}{7}$

Answer 1

The correct option is D $\frac{3}{7}$

Let $A_1$ be the event $\le 3$ (when a die is cast) $A_2$ be the event $\ge 4$ (when a die is cast) ${}^{\prime }{E}^{\prime }$ be the event drawn ball is Black.
$\therefore P\left(A_1\right)=\frac{1}{2}=P\left(A_2\right)$
$P\left(\frac{E}{A_1}\right)=\frac{3}{7},P\left(\frac{E}{A_2}\right)=\frac{4}{7}$
$\therefore P\left(\frac{A_1}{E}\right)=\frac{P\left(A_1\right)P\left(\frac{E}{A_1}\right)}{P\left(A_1\right)P\left(\frac{E}{A_1}\right)+P\left(A_2\right)P\left(\frac{E}{A_2}\right)}$
$=\frac{\frac{1}{2}\times \frac{3}{7}}{\frac{1}{2}\times \frac{3}{7}+\frac{1}{2}\times \frac{4}{7}}=\frac{3}{7}$
Hence choice (d) is correct.