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

• A. $\frac { 7 } { 15 }$
• B. $\frac { 5 } { 19 }$
• C. $\frac { 3 } { 4 }$
• D. None of these

Answer 1

Answer: (A)

$\frac { 7 } { 15 }$

Answer 2

Consider the following events:

$A \rightarrow$ Ball drawn is black; $E _ { 1 } \rightarrow$ Bag I is chosen;

$E _ { 2 } \rightarrow$ Bag II is chosen and $E _ { 3 } \rightarrow$ Bag III is chosen.

Then $P \left( E _ { 1 } \right) = \left( E _ { 2 } \right) = P \left( E _ { 3 } \right) = \frac { 1 } { 3 } , P \left( \frac { A } { E _ { 1 } } \right) = \frac { 3 } { 5 }$.

$P \left( \frac { A } { E _ { 2 } } \right) = \frac { 1 } { 5 } , P \left( \frac { A } { E _ { 3 } } \right) = \frac { 7 } { 10 }$

Required probability $= P \left( \frac { E _ { 3 } } { A } \right)$

$= \frac { P \left( E _ { 3 } \right) P \left( A / E _ { 3 } \right) } { P \left( E _ { 1 } \right) P \left( A / E _ { 1 } \right) + P \left( E _ { 2 } \right) P \left( A / E _ { 2 } \right) + P \left( E _ { 3 } \right) P \left( A / E _ { 3 } \right) } = \frac { 7 } { 15 }$ .