Probability Question 183
Question: There are 3 bags which are known to contain 2 white and 3 black balls; 4 white and 1 black balls and 3 white and 7 black balls respectively. A ball is drawn at random from one of the bags and found to be a black ball. Then the probability that it was drawn from the bag containing the most black balls is
Options:
A) $ \frac{7}{15} $
B) $ \frac{5}{19} $
C) $ \frac{3}{4} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Consider the following events :
$ A\to $ Ball drawn is black;
$ E_1\to $ Bag I is chosen;
$ E_2\to $ Bag II is chosen and $ E_3\to $ Bag III is chosen.
Then $ P(E_1)=(E_2)=P(E_3)=\frac{1}{3},P( \frac{A}{E_1} )=\frac{3}{5}. $
$ P( \frac{A}{E_2} )=\frac{1}{5},P( \frac{A}{E_3} )=\frac{7}{10} $
Required probability $ =P( \frac{E_3}{A} ) $
$ =\frac{P(E_3)P(A/E_3)}{P(E_1)P(A/E_1)+P(E_2)P(A/E_2)+P(E_3)P(A/E_3)}=\frac{7}{15} $ .
BETA
