Q108 / 170 · Manipal (MET) 2015 · Mathematics
Let $u_1$ and $u_2$ be two urns such that $u_1$ contains 3 white, 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 urn $u_1$ 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 $u_2$. Then, probability of the drawn ball from $u_2$ being white is
- A$\frac{13}{30}$
- Correct answerB$\frac{23}{30}$
- C$\frac{19}{30}$
- D$\frac{11}{30}$
Solution
By total probability theorem,
Required probability $=\frac{1}{2} \times\left(\frac{3}{5} \times 1+\frac{2}{5} \times \frac{1}{2}\right)$
$+\frac{1}{2}\left(\frac{{ }^3 C_2}{{ }^5 C_2} \times 1+\frac{{ }^2 C_2}{{ }^5 C_2} \times \frac{1}{3}+\frac{{ }^3 C_1 \times{ }^2 C_1}{{ }^5 C_2} \times \frac{2}{3}\right)$
$\begin{aligned} & =\frac{1}{2}\left(\frac{3}{5}+\frac{1}{5}\right)+\frac{1}{2}\left(\frac{3}{10}+\frac{1}{30}+\frac{2}{5}\right) \\ & =\frac{2}{5}+\frac{1}{2}\left(\frac{9+1+12}{30}\right) \\ & =\frac{2}{3}+\frac{22}{60}=\frac{24+22}{60} \\ & =\frac{46}{60}=\frac{23}{30}\end{aligned}$
Hence, the correct answer is option: 'B'.