Probability Question 336
Question: For k=1, 2, 3 the box $ B_{k} $ contains k red balls and $ (k+1) $ white balls. Let $ P(B_1)=\frac{1}{2},P(B_2)=\frac{1}{3} $ and $ P(B_3)=\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
Options:
A) $ \frac{35}{78} $
B) $ \frac{14}{39} $
C) $ \frac{10}{13} $
D) $ \frac{12}{13} $
Show Answer
Answer:
Correct Answer: B
Solution:
In a box, $ B_1=1R,2W;B_2=2R,3W $ and $ B_3=3R,4W $
Also, given that, $ P(B_1)=\frac{1}{2},P(B_2)=\frac{1}{3} $ and $ P(B_3)=\frac{1}{6} $
$ \therefore P( \frac{B_2}{R} ) $
$ =\frac{P(B_2)P( \frac{R}{B_2} )}{P(B_1)P( \frac{R}{B_1} )+P(B_2)P( \frac{R}{B_2} )+P(B_3)P( \frac{R}{B_3} )} $
$ =\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}. $
BETA
