Statistics And Probability Question 459
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} $
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Answer:
Correct Answer: D
Solution:
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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
