Probability Question 197
Question: A bag A contains 2 white and 3 red balls and bag B contains 4 white and 5 red balls. One ball is drawn at random from a randomly chosen bag and is found to be red. The probability that it was drawn from bag B was
[BIT Ranchi 1988; IIT 1976]
Options:
A) $ \frac{5}{14} $
B) $ \frac{5}{16} $
C) $ \frac{5}{18} $
D) $ \frac{25}{52} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ E_1 $ be the event that the ball is drawn from bag $ A,E_2 $ the event that it is drawn from bag $ B $ and $ E $ that the ball is red.We have to find $ P(E_2/E) $ .
Since both the bags are equally likely to be selected, we have $ P(E_1)=P(E_2)=\frac{1}{2} $
Also $ P(E/E_1)=3/5 $ and $ P(E/E_2)=5/9. $
Hence by Bay’s theorem, we have $ P(E_2/E)=\frac{P(E_2)P(E/E_2)}{P(E_1)P(E/E_1)+P(E_2)P(E/E_2)} $ $ =\frac{\frac{1}{2}.\frac{5}{9}}{\frac{1}{2}.\frac{3}{5}+\frac{1}{2}.\frac{5}{9}}=\frac{25}{52}. $
BETA
