Paragraph: Let $U_1$ and $U_2$ be two urns such that $U_1$ contains 3 white and 2 red balls and $U_2$…

Paragraph: 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 $U_2$. Question: The probability of the drawn ball from $U_2$ being white is
  1. $\frac{13}{30}$
  2. $\frac{23}{30}$
  3. $\frac{19}{30}$
  4. $\frac{11}{30}$

Solution

Now, probability of the drawn ball from $\mathrm{V}_2$ being white is $ \Rightarrow P \text { (white/ } V_2 \text { ) } $ $ \begin{aligned} & \left.\Rightarrow P \text { (white/ } V_2\right) \\ & =P(H) \cdot\left\{\frac{{ }^3 C_1}{{ }^5 C_1} \times \frac{{ }^2 C_1}{{ }^2 C_1}+\frac{{ }^2 C_1}{{ }^5 C_1} \times \frac{{ }^1 C_1}{{ }^2 C_1}\right\} \\ & +P(T)\left\{\frac{{ }^3 C_2}{{ }^5 C_2} \times \frac{{ }^3 C_2}{{ }^3 C_2}+\frac{{ }^2 C_2}{{ }^5 C_2} \times \frac{{ }^1 C_1}{{ }^3 C_2}\right. \\ & \left.+\frac{{ }^3 C_1 \cdot{ }^2 C_1}{{ }^5 C_2} \times \frac{{ }^2 C_1}{{ }^3 C_2}\right\} \\ & \text { Now, } P\left(\text { white } / V_2\right)=\frac{1}{2}\left\{\frac{3}{5} \times 1+\frac{2}{5} \times \frac{1}{2}\right\} \\ & +\frac{1}{2}\left\{\frac{3}{10} \times 1+\frac{1}{10} \times \frac{1}{3}+\frac{6}{10} \times \frac{2}{3}\right\}=\frac{23}{30} \\ & \end{aligned} $

Asked in: JEE Advanced 2011 (Paper 1)

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