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 B is

• A. $\frac { 5 } { 14 }$
• B. $\frac { 5 } { 16 }$
• C. $\frac { 5 } { 18 }$
• D. $\frac { 25 } { 52 }$

Answer 1

Answer: (D)

$\frac { 25 } { 52 }$

Answer 2

Let E₁ be the event that the ball is drawn from bag A, E₂ the event that it is drawn from bag B and E that the ball is red.

We have to find $P \left( E _ { 2 } / E \right)$ .

Since both the bags are equally likely to be selected,

we have $P \left( E _ { 1 } \right) = P \left( E _ { 2 } \right) = \frac { 1 } { 2 }$ . Also $P \left( E / E _ { 1 } \right) = 3 / 5$ and

$P \left( E / E _ { 2 } \right) = 5 / 9$.

Hence by Baye’s theorem, we have

.