Q. If from each of three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn, is
A) $\frac{13}{32}$
B) $\frac{1}{4}$
C) $\frac{1}{32}$
D) $\frac{3}{16}$
Explanation
(a)
$P$(2 white and 1 black)
$= P(W_{1}W_{2}B_{3}\ or\ W_{1}B_{2}W_{3}\ or\ B_{1}W_{2}W_{3})$
$= P\left( W_{1} \right)P\left( W_{2} \right)P\left( B_{3} \right) + P\left( W_{1} \right)P\left( B_{2} \right)P\left( W_{3} \right) + P\left( B_{1} \right)P\left( W_{2} \right)P(W_{3})$
$= \left( \frac{3}{4} \right)\left( \frac{2}{4} \right)\left( \frac{3}{4} \right) + \left( \frac{3}{4} \right)\left( \frac{2}{4} \right)\left( \frac{1}{4} \right) + \left( \frac{1}{4} \right)\left( \frac{2}{4} \right)\left( \frac{1}{4} \right)$
$= \frac{1}{32}(9 + 3 + 1) = \frac{13}{32}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.