Probability Ques 53
If from each of the 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 balls will be drawn, is
(a) $\frac{13}{32}$
(b) $\frac{1}{4}$
(c) $\frac{1}{32}$
(d) $\frac{3}{16}$
(1998, 2M)
Show Answer
Answer:
Correct Answer: 53.(a)
Solution:
Formula:
- $P(2$ white and 1 black $)=P\left(W _1 W _2 B _3\right.$ or $W _1 B _2 W _3$ or $ \left.B _1 W _2 W _3\right) $
$= P\left(W _1 W _2 B _3\right)+P\left(W _1 B _2 W _3\right)+P\left(B _1 W _2 W _3\right) $
$= 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\left(W _3\right) $
$=\frac{3}{4} \cdot \frac{2}{4} \cdot \frac{3}{4}+\frac{3}{4} \cdot \frac{2}{4} \cdot \frac{1}{4}+\frac{1}{4} \cdot \frac{2}{4} \cdot \frac{1}{4}= \frac{1}{32}(9+3+1)=\frac{13}{32}$
BETA
