Probability Question 199
Question: Bag A contains 4 green and 3 red balls and bag B contains 4 red and 3 green balls. One bag is taken at random and a ball is drawn and noted it is green. The probability that it comes bag B
[DCE 2005]
Options:
A) $ \frac{2}{7} $
B) $ \frac{2}{3} $
C) $ \frac{3}{7} $
D) $ \frac{1}{3} $
Show Answer
Answer:
Correct Answer: C
Solution:
It is based on Baye’s theorem.
Probability of picked bag A $ P(A)=\frac{1}{2} $
Probability of picked bag B $ P(B)=\frac{1}{2} $
Probability of green ball picked from bag A $ =P(A).P( \frac{G}{A} ) $
$ =\frac{1}{2}\times \frac{4}{7}=\frac{2}{7} $
Probability of green ball picked from bag B $ =P(B).P( \frac{G}{B} ) $
$ =\frac{1}{2}\times \frac{3}{7}=\frac{3}{14} $ Total probability of green ball = $ \frac{2}{7}+\frac{3}{14}=\frac{1}{2} $
Probability of fact that green ball is drawn from bag B
$ P( \frac{G}{B} )=\frac{P(B)P( \frac{G}{B} )}{P(A)P( \frac{G}{A} )+P(B)P( \frac{G}{B} )}=\frac{\frac{1}{2}\times \frac{3}{7}}{\frac{1}{2}\times \frac{4}{7}+\frac{1}{2}\times \frac{3}{7}}=\frac{3}{7} $ .
BETA
