Question

For $k=1,2,3$ the box $Bk$ contains $k$ red balls and $(k+1)$ white balls. Let $P\left(B_{1}\right)=\frac{1}{2}, P\left(B_{2}\right)=\frac{1}{3}=\frac{1}{3}$ 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

Answer: (B)

$\frac{14}{39}$

Answer 2

In a box, $B_{1}=1 R, 2 W ; B_{2}=2 R, 3 W$
and $B_{3}=3 R, 4 W$ Also, given that, $P\left(B_{1}\right)=\frac{1}{2}, P\left(B_{2}\right)$
$=\frac{1}{3} P\left(B_{3}\right)=\frac{1}{6}$
$\therefore P\left(\frac{B_{2}}{R}\right)=\frac{P\left(B_{2}\right) P\left(\frac{R}{B_{2}}\right)}{P\left(B_{1}\right) P\left(\frac{R}{B_{1}}\right)+P\left(B_{2}\right) P\left(\frac{R}{B_{2}}\right)+P\left(B_{3}\right) P\left(\frac{R}{B_{3}}\right)}$
$=\frac{\frac{1}{3} \times \frac{2}{5}}{\frac{1}{2} \times \frac{1}{3}+\frac{1}{3} \times \frac{2}{5}+\frac{1}{6} \times \frac{3}{7}}$
$=\frac{\frac{2}{15}}{\frac{1}{6}+\frac{2}{15}+\frac{1}{14}}$
$=\frac{14}{39}$