Conditional Probability of Head Given Drawn Ball is White

To find the probability that head appeared on the coin given that the drawn ball from $U_2$ is white, we use Bayes' Theorem.

Let:

• $H$ be the event that a Head appears on the coin.
• $T$ be the event that a Tail appears on the coin.
• $E$ be the event that the ball drawn from urn $U_2$ is white.

Since the coin is fair:
$P(H) = \frac{1}{2}, \quad P(T) = \frac{1}{2}$

Urn $U_1$ contains $3$ white balls and $2$ red balls (total $5$ balls).
Initially, urn $U_2$ contains $1$ white ball.

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Step 1: Probability of drawing a white ball from $U_2$ given Head appeared, $P(E \mid H)$

If Head appears, $1$ ball is drawn from $U_1$ and transferred to $U_2$.

• The ball drawn from $U_1$ is white with probability $\frac{3}{5}$. In this case, $U_2$ now has $2$ white balls out of $2$ balls. The probability of drawing a white ball from $U_2$ is $\frac{2}{2} = 1$.
• The ball drawn from $U_1$ is red with probability $\frac{2}{5}$. In this case, $U_2$ now has $1$ white ball and $1$ red ball out of $2$ balls. The probability of drawing a white ball from $U_2$ is $\frac{1}{2}$.

By the Law of Total Probability:
$P(E \mid H) = \left(\frac{3}{5} \times 1\right) + \left(\frac{2}{5} \times \frac{1}{2}\right) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$

Thus,
$P(H \cap E) = P(H) \times P(E \mid H) = \frac{1}{2} \times \frac{4}{5} = \frac{4}{10} = \frac{12}{30}$

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Step 2: Probability of drawing a white ball from $U_2$ given Tail appeared, $P(E \mid T)$

If Tail appears, $2$ balls are drawn from $U_1$ and transferred to $U_2$. The total number of ways to choose $2$ balls from $5$ is $\binom{5}{2} = 10$.

• Both transferred balls are white:
  $\text{Probability} = \frac{\binom{3}{2}}{\binom{5}{2}} = \frac{3}{10}$
  Now, $U_2$ contains $1 + 2 = 3$ white balls and $0$ red balls (total $3$ balls).
  The probability of drawing a white ball from $U_2$ is $\frac{3}{3} = 1$.

• One white and one red ball are transferred:
  $\text{Probability} = \frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{3 \times 2}{10} = \frac{6}{10}$
  Now, $U_2$ contains $1 + 1 = 2$ white balls and $1$ red ball (total $3$ balls).
  The probability of drawing a white ball from $U_2$ is $\frac{2}{3}$.

• Both transferred balls are red:
  $\text{Probability} = \frac{\binom{2}{2}}{\binom{5}{2}} = \frac{1}{10}$
  Now, $U_2$ contains $1$ white ball and $2$ red balls (total $3$ balls).
  The probability of drawing a white ball from $U_2$ is $\frac{1}{3}$.

Thus:
$P(E \mid T) = \left(\frac{3}{10} \times 1\right) + \left(\frac{6}{10} \times \frac{2}{3}\right) + \left(\frac{1}{10} \times \frac{1}{3}\right) = \frac{3}{10} + \frac{4}{10} + \frac{1}{30} = \frac{7}{10} + \frac{1}{30} = \frac{22}{30} = \frac{11}{15}$

Hence,
$P(T \cap E) = P(T) \times P(E \mid T) = \frac{1}{2} \times \frac{11}{15} = \frac{11}{30}$

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Step 3: Total Probability of event $E$

$P(E) = P(H \cap E) + P(T \cap E) = \frac{12}{30} + \frac{11}{30} = \frac{23}{30}$

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Step 4: Conditional Probability $P(H \mid E)$

Applying Bayes' Theorem:
$P(H \mid E) = \frac{P(H \cap E)}{P(E)} = \frac{\frac{12}{30}}{\frac{23}{30}} = \frac{12}{23}$

Therefore, the correct option is D.