JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Probability XII

Let

U_{1}

and

U_{2}

be two urns such that

U_{1}

contains 3 white and 2 red balls, and

U_{2}

contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from

U_{1}

and put into

U_{2}

. However, if tail appears then 2 balls are drawn at random from

U_{1}

and put into

U_{2}

. Now 1 ball is drawn at random from

U_{2}

.

Given that the drawn ball from

U_{2}

is white, the probability that head appeared on the coin is

Answer: (D)

Solution

Using Bayes' theorem:

P (H ∣ W) = \frac{P ( W ∣ H ) \cdot P ( H )}{P ( W )}

.

From the previous question,

P (W) = \frac{23}{30}

P (W ∣ H) = \frac{3}{5} \cdot 1 + \frac{2}{5} \cdot \frac{1}{2} = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}

P (H) = \frac{1}{2}

.

Therefore:

P (H ∣ W) = \frac{\frac{4}{5} \cdot \frac{1}{2}}{\frac{23}{30}} = \frac{\frac{2}{5}}{\frac{23}{30}} = \frac{2}{5} \cdot \frac{30}{23} = \frac{60}{115} = \frac{12}{23}

JEE Advanced · 2011 · 10 Apr 2011 · Paper 1 · Single correct option