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}

.

The probability of the drawn ball from

U_{2}

being white is

Answer: (B)

Solution

Using the law of total probability:

Case 1: Head (1 ball transferred)

P(H) = 1/2
P(white from $U_{1}$ ) = 3/5 → $U_{2}$ has 2 white → P(white from $U_{2}$ ) = 1
P(red from $U_{1}$ ) = 2/5 → $U_{2}$ has 1 white, 1 red → P(white from $U_{2}$ ) = 1/2

Case 2: Tail (2 balls transferred)

P(T) = 1/2
P(2 white) = $\frac{( 2 3 )}{( 2 5 )} = \frac{3}{10}$ → $U_{2}$ has 3 white → P(white) = 1
P(2 red) = $\frac{( 2 2 )}{( 2 5 )} = \frac{1}{10}$ → $U_{2}$ has 1 white, 2 red → P(white) = 1/3
P(1 white, 1 red) = $\frac{3 \cdot 2}{10} = \frac{6}{10}$ → $U_{2}$ has 2 white, 1 red → P(white) = 2/3

Total probability:

P (white) = \frac{1}{2} (\frac{3}{5} \cdot 1 + \frac{2}{5} \cdot \frac{1}{2}) + \frac{1}{2} (\frac{3}{10} \cdot 1 + \frac{1}{10} \cdot \frac{1}{3} + \frac{6}{10} \cdot \frac{2}{3})

= \frac{1}{2} (\frac{3}{5} + \frac{1}{5}) + \frac{1}{2} (\frac{3}{10} + \frac{1}{30} + \frac{4}{10})

= \frac{1}{2} \cdot \frac{4}{5} + \frac{1}{2} (\frac{9 + 1 + 12}{30}) = \frac{2}{5} + \frac{11}{30} = \frac{12 + 11}{30} = \frac{23}{30}

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