Question

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 the head appears then 1 ball is drawn at random from $U_1$and put into $U_2$. However, if the 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 a head appeared on the coin is

• A. $\frac{17}{23}$
• B. $\frac{11}{23}$
• C. $\frac{15}{23}$
• D. $\frac{12}{23}$

Answer 1

The correct option is D

$\frac{12}{23}$

Explanation for Correct Answer:

Given: Urn $U_1$contains $3$ white and $2$ red balls .

$\Rightarrow U_1=2+3=5$

Urn $U_2$ contains only $1$ white ball

$\Rightarrow U_2=1$

Let $W$ be the event of drawing white ball.

Then, the probability of drawing white balls from $U_2$, when head appears on coin.

$$\begin{gathered} \Rightarrow P_1=\frac{1}{2}\times \left[\frac{3}{5}\times \frac{2}{2}+\frac{2}{5}\times \frac{1}{2}\right] \\ =\frac{1}{2}\times \left[\frac{3}{5}\times 1+\frac{1}{5}\right] \\ =\frac{1}{2}\times \left[\frac{3+1}{5}\right] \\ =\frac{1}{2}\times \left[\frac{4}{5}\right] \\ =\frac{2}{5} \end{gathered}$$

Now, the probability of drawing white balls from $U_2$, when tail appears on coin.

$$\begin{gathered} \Rightarrow P_2=\frac{1}{2}\times \left[\frac{3}{5}\times \frac{2}{2}+\frac{2}{5}\times \frac{1}{2}\right]+\frac{2}{5}\left\{\frac{1}{2}\times \left[\frac{3}{10}\times 1+\frac{1}{10}\times \frac{1}{3}+\frac{6}{10}\times \frac{2}{3}\right]\right\} \\ =\frac{1}{2}\times \left[\frac{3}{5}+\frac{1}{5}\right]+\frac{2}{5}\left[\frac{1}{2}\times \left(\frac{9+1+12}{30}\right)\right] \\ =\frac{1}{2}\times \left[\frac{4}{5}\right]+\frac{2}{5}\left[\frac{1}{2}\times \left(\frac{22}{30}\right)\right] \\ =\frac{1}{2}\times \left[\frac{4}{5}\right]+\frac{1}{5}\left(\frac{11}{15}\right) \\ =\frac{2}{5}+\frac{11}{75} \\ =\frac{23}{30} \end{gathered}$$

Now, to find the probability that head appeared on the coin,

$$\begin{gathered} \Rightarrow P=\frac{P_1}{P_2} \\ =\frac{{\displaystyle \frac{2}{5}}}{{\displaystyle \frac{23}{30}}} \\ =\frac{2}{5}\times \frac{30}{23} \\ =\frac{12}{23} \end{gathered}$$

Therefore, the probability that head appeared on the coin is $\frac{12}{23}$

Hence, option (D) is the correct answer