Question

For $k=1,2,3$ the box $B_k$ contains $k$ red balls and $(k+1)$ white balls, let $P\left(B_1\right)=\frac{1}{2},P\left(B_2\right)=1$ and $P\left(B_3\right)=\frac{1}{6}$. A box is selected at random and a ball is drawn from it. If a red ball is drawn, then the probability that it has come from box $B_2$ is

• A. $\frac{35}{78}$
• B. $\frac{14}{39}$
• C. $\frac{10}{13}$
• D. $\frac{12}{13}$

Answer 1

The correct option is B $\frac{14}{39}$

In a box,
$B_1=1R,2W$
$B_2=2R,3W$
and $B_3=3R,4W$

Also, given that,
$P\left(B_1\right)=\frac{1}{2},P\left(B_2\right)=\frac{1}{3}$ and $P\left(B_3\right)=\frac{1}{6}$

$\therefore P\left(\frac{B_2}{R}\right)=\frac{P\left(B_2\right)P\left(\frac{R}{B_2}\right)}{P\left(B_1\right)P\left(\frac{R}{B_1}\right)+P\left(B_2\right)P\left(\frac{R}{B_2}\right)+P\left(B_3\right)P\left(\frac{R}{B_3}\right)}$

$=\frac{\frac{1}{3}\times \frac{2}{5}}{\frac{1}{2}\times \frac{1}{3}\times \frac{1}{3}\times \frac{2}{5}+\frac{1}{6}\times \frac{3}{7}}=\frac{\frac{2}{15}}{\frac{1}{6}+\frac{2}{15}+\frac{1}{14}}$

$=\frac{\frac{2}{15}}{\frac{35+28+15}{210}}=\frac{2}{15}\times \frac{210}{78}=\frac{14}{39}$