Q. A bag $X$ contains 2 white and 3 black balls and another bag $Y$ contains 4 white and 2 black balls. One bag is selected at random and a ball is drawn from it. Then, probability for the ball chosen be white, is
A) $\frac{2}{25}$
B) $\frac{7}{15}$
C) $\frac{8}{15}$
D) $\frac{14}{15}$
Explanation
(c)
Let $A$ be the event of selecting bag $X,\ B$ be the event of selecting bag $Y$ and $E$ be the event of drawing a white ball, then probability of selecting a bag
$P(A) = \frac{1}{2},\ P(B) = \frac{1}{2},$
$P\left( \frac{E}{A} \right) = \frac{2}{5},\ P\left( \frac{E}{B} \right) = \frac{4}{6} = \frac{2}{3}$.
Required probability,
$P(E) = P(A)\ P\left( \frac{E}{A} \right) + P(B)\ P\left( \frac{E}{B} \right)$
$= \frac{1}{2} \bullet \frac{2}{5} + \frac{1}{2} \bullet \frac{2}{3}$
$= \frac{8}{15}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.