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 $U_2$. The probability of the drawn ball from $U_2$ being white is

The correct option is B

$\frac{23}{30}$

Explanation for the correct options:

Find the required probability:

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

$U_2$ contains only $1$ white ball

By using the given condition that is ,

Step 1: The probability of drawing ball from $U_1$, if the ball drawn at random is white.

$$\begin{gathered} \Rightarrow P_1=\frac{1}{2}\times \frac{3}{5}\times \frac{2}{2} \\ =\frac{1\times 3}{1\times 5} \\ =\frac{3}{5} \end{gathered}$$

Similarly,

Step 2 : The probability of drawing ball from $U_2$, if the ball drawn at random is red.

$$\begin{gathered} \Rightarrow P_2=\frac{1}{2}\times \frac{2}{5}\times \frac{1}{2} \\ =\frac{1}{2\times 5} \\ =\frac{1}{10} \end{gathered}$$

Also using the condition of a coin tossed that is ,

Step 3: The probability of drawing ball from $U_1$, if the ball drawn at random is two white.

$$\begin{gathered} \Rightarrow P_3=\frac{1}{2}\times \frac{{}^3{\mathrm{ℂ}}_2}{{}^5{\mathrm{ℂ}}_2}\times \frac{3}{3} \\ =\frac{1}{2}\times \frac{{\displaystyle \frac{3!}{2!1!}}}{\frac{5!}{2!3!}}\times 1 \\ =\frac{1}{2}\times \frac{3!\times 2!}{2!}\times \frac{3!\times 2\times 1}{5\times 4\times 3!} \\ =\frac{1}{2}\times 3\times \frac{2}{5\times 4} \\ =\frac{3}{5\times 4} \\ =\frac{3}{20} \end{gathered}$$

Step 4: The probability of drawing ball from $U_2$, if the ball drawn at random is$1$ red and $1$ white ball.

$$\begin{gathered} \Rightarrow P_4=\frac{1}{2}\times \frac{{}^3{\mathrm{ℂ}}_1{\times }^2{\mathrm{ℂ}}_1}{{}^5{\mathrm{ℂ}}_2}\times \frac{2}{3} \\ =\frac{1}{2}\times \frac{{\displaystyle \frac{3!}{2!1!}\times \frac{2!}{1!1!}}}{\frac{5!}{2!3!}}\times \frac{2}{3} \\ =\frac{1}{2}\times \frac{3!\times 2!}{2!}\times 2\times \frac{3!\times 2\times 1}{5\times 4\times 3!} \\ =\frac{2\times 2}{5\times 4} \\ =\frac{4}{5\times 4} \\ =\frac{1}{5} \end{gathered}$$

Step 5: Also, the probability of drawing a ball from $U_2$. If the ball drawn at random $2$ is red ball

$$\begin{gathered} \Rightarrow P_5=\frac{1}{2}\times \frac{{}^2{\mathrm{ℂ}}_2}{{}^5{\mathrm{ℂ}}_2}\times \frac{1}{3} \\ =\frac{1}{2}\times \frac{{\displaystyle \frac{2!}{2!1!}}}{\frac{5!}{2!3!}}\times \frac{2}{3} \\ =\frac{1}{2}\times 1\times \frac{3!\times 2!}{5\times 4\times 3!}\times \frac{1}{3} \\ =\frac{1}{5\times 4\times 3} \\ =\frac{1}{60} \end{gathered}$$

So, required probability is $\frac{3}{10}+\frac{1}{10}+\frac{3}{20}+\frac{1}{5}+\frac{1}{60}=\frac{23}{30}$

Hence, option (B) is the correct answer