Q. For $k = 1,2,3$ the box $B_{k}$ contains $k$ red balls and $(k + 1)$ white balls, Let $P\left( B_{1} \right) = \frac{1}{2},P\left( B_{2} \right) = \frac{1}{3}$ 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) $\frac{12}{13}$
Explanation
(b)
$P\left( \frac{B_{2}}{R} \right) = \frac{P\left( B_{2} \right)P\left( \frac{R}{B_{2}} \right)}{P\left( B_{1} \right)P\left( \frac{R}{B_{1}} \right) + P\left( B_{2} \right)P\left( \frac{R}{B_{2}} \right) + P\left( B_{3} \right)P\left( \frac{R}{B_{3}} \right)}$
$= \frac{\frac{1}{3} \times \frac{2}{5}}{\frac{1}{2} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{5} + \frac{1}{6} \times \frac{3}{7}}$
$= \frac{\frac{2}{15}}{\frac{1}{6} + \frac{2}{15} + \frac{1}{14}} = \frac{14}{39}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.