Three boxes $\mathrm{B}_1, \mathrm{~B}_2$ and $\mathrm{B}_3$ contain balls with different colors as follows:…
Three boxes $\mathrm{B}_1, \mathrm{~B}_2$ and $\mathrm{B}_3$ contain balls with different colors as follows:
\begin{array}{cccc}
& White & Black & Red \\
\mathrm{B}_1 & 2 & 1 & 2 \\
\mathrm{~B}_2 & 3 & 2 & 4 \\
\mathrm{~B}_3 & 4 & 3 & 2
\end{array}
A die is thrown. Box $B_1$ is chosen if either 1 or 2 turns up. Box $B_2$ is chosen if 3 or 4 turns up and box $B_3$ is chosen if 5 or 6 turns up. Having chosen a box in this way, a ball is drawn at random from that box. If the ball drawn is found to be Red, then the probability that it is drawn from box $\mathrm{B}_2$ is
$\frac{7}{12}$
$\frac{5}{12}$
$\frac{1}{12}$
$\frac{3}{26}$
Solution
If 1 or 2 turns up, box $B_1$ is selected
Hence, $P\left(B_1\right)=\frac{2}{6}$
Similary, $\mathrm{P}\left(\mathrm{B}_2\right)=\frac{2}{6}$ and $\mathrm{P}\left(\mathrm{B}_3\right)=\frac{2}{6}$
For box $\mathrm{B}_1$ :
Total number of balls $=5$
No. of red balls $=2$
$\therefore$ Probability of red ball $=\frac{2}{5} \Rightarrow P\left(\frac{R}{B_1}\right)=\frac{2}{5}$
Similarly, $\mathrm{P}\left(\frac{\mathrm{R}}{\mathrm{B}_2}\right)=\frac{4}{9}$ and $\mathrm{P}\left(\frac{\mathrm{R}}{\mathrm{B}_3}\right)=\frac{2}{9}$ Now, the required probability is
$P\left(\frac{B_2}{R}\right)=\frac{P\left(\frac{R}{B_2}\right) P\left(B_2\right)}{P\left(\frac{R}{B_1}\right) P\left(B_1\right)+P\left(\frac{R}{B_2}\right) P\left(B_2\right)+P\left(\frac{R}{B_3}\right) P\left(B_3\right)}$
$=\frac{\frac{4}{9} \times \frac{2}{6}}{\frac{2}{6} \times \frac{2}{5}+\frac{4}{9} \times \frac{2}{6}+\frac{2}{9} \times \frac{2}{6}}=\frac{5}{12}$