Question

Bag A contains 3W and 4B balls. Bag B contains 2W & 3B balls. A ball is transferred from A to B. Then the probability of drawing a white ball from Bag B is

• A. 17/42
• B. 18/42
• C. 7/42
• D. 9/42

Answer 1

Answer: (A)

17/42

Answer 2

Let A and B events of transferring white and Black balls respectively.

P(1) = $\frac { 3 } { 7 }$ P(2) = $\frac { 4 } { 7 }$

Let E be the event of drawing white ball for Bag B

P(E/A) = $\frac { 3 } { 6 } = \frac { 1 } { 2 }$; P(E/B) = $\frac { 2 } { 6 } = \frac { 1 } { 3 }$

P(5) = P(1). P(E/A) + P(2). P(E/B)

= $\frac { 3 } { 7 } \times \frac { 1 } { 2 } + \frac { 4 } { 7 } \times \frac { 1 } { 3 } = \frac { 9 + 8 } { 7 \times 6 } + \frac { 17 } { 42 }$