Question

Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains only 1 white ball. A fair coin is tossed. If the head appears then 1 ball is drawn at random from U1 and put into U2. However, if tail appears then 2 balls are drawn at random from U1 and put into U2. Now 1 ball is drawn at random from U2. Given that the drawn ball from U2 is white, the probability that a head appeared on the coin is (a) \(\frac{17}{23}\) (b)\(\frac{11}{23}\) (c) \(\frac{15}{23}\) (d) \(\frac{12}{23}\) Select the correct answer from above options

Answer 1

Answer : (D) = \(\frac{12}{23}\) The movement of balls from Urn 1 to Urn 2 on the condition that head or tail appears on the coin can be shown as: Let the events be defined as: W : Drawing a white ball from U2 H : Tossing a head T : Tossing a tail. \(\therefore\) P(H) = P(T) = \(\frac{1}{2}\) P(W/H) = Probability of drawing a white ball from Urn 2, when head is tossed \(= \frac{^3C_1}{^5C_1} \times \frac{^2C_1}{^2C_1} +\frac{^2C_1}{^5C_1} \times \frac{^1C_1}{^2C_1}\) From(i) = \(\frac{3}{5} \times 1 +\frac{2}{5} \times \frac{1}{2} = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}\) P(W/T) = Probability of drawing a white ball from Urn 2, when tail is tossed \(= \frac{^3C_2}{^5C_2} \times \frac{^3C_1}{^3C_2} +\frac{^2C_2}{^5C_2} \times \frac{^1C_1}{^3C_1} + \frac{^3C_1 \times ^2C_1}{^5C_2} \times \frac{^2C_1}{^3C_2}\) (see diag.(iii)) = \(\frac{3\times 2}{5 \times 4} \times 1 +\frac{1\times 2}{5 \times 4}\times \frac{1}{3} + \frac{3\times 2\times 2}{5\times 4} \times \frac{2}{3}\) = \(\frac{3}{10}+\frac{1}{30}+\frac{2}{5} = \frac{9+1+12}{30} = \frac{22}{30}\) \(P\big(\frac{H}{W}\big) = \frac{P(H).P(W/H)}{P(H).P(W/H) + P(T).P(W/T)}\) = \(\frac{\frac{1}{2} \times \frac{4}{5}}{\frac{1}{2}\times \frac{4}{5} +\frac{1}{2} \times \frac{22}{30}}\) = \(\frac{4/5}{4/5 +22/30} = \frac{4/5}{\frac {24+22}{30}}\) = \(\frac{4/5}{46/30} = \frac{4}{5}\times \frac{30}{46}\) = \(\frac{12}{23}\)