Probability of Drawing White Ball from Urn Two

To find the probability that the ball drawn from $U_2$ is white, we use the Law of Total Probability.

Let $H$ denote the event that head appears on tossing the fair coin, and $T$ denote the event that tail appears.
$P(H) = \frac{1}{2}, \quad P(T) = \frac{1}{2}$

Let $W$ be the event that the ball drawn from $U_2$ is white.

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Case 1: Head appears ($H$)

One ball is drawn at random from $U_1$ (which contains $3$ white and $2$ red balls) and put into $U_2$ (which initially contains $1$ white ball).

• Subcase 1.1: A white ball is transferred from $U_1$ to $U_2$.

  • Probability of transferring a white ball: $P(W_1 | H) = \frac{3}{5}$.
  • Urn $U_2$ now contains $2$ white balls and $0$ red balls.
  • Probability of drawing a white ball from $U_2$: $1$.

• Subcase 1.2: A red ball is transferred from $U_1$ to $U_2$.

  • Probability of transferring a red ball: $P(R_1 | H) = \frac{2}{5}$.
  • Urn $U_2$ now contains $1$ white ball and $1$ red ball.
  • Probability of drawing a white ball from $U_2$: $\frac{1}{2}$.

Thus, the conditional probability $P(W | H)$ is:
$P(W | H) = \left(\frac{3}{5} \times 1\right) + \left(\frac{2}{5} \times \frac{1}{2}\right) = \frac{3}{5} + \frac{1}{5} = \frac{4}{5}$

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Case 2: Tail appears ($T$)

Two balls are drawn at random from $U_1$ and put into $U_2$. The total number of ways to pick $2$ balls out of $5$ is $\binom{5}{2} = 10$.

• Subcase 2.1: Two white balls are transferred.

  • Probability: $\frac{\binom{3}{2}}{\binom{5}{2}} = \frac{3}{10}$.
  • Urn $U_2$ now contains $3$ white balls and $0$ red balls.
  • Probability of drawing a white ball from $U_2$: $1$.

• Subcase 2.2: One white and one red ball are transferred.

  • Probability: $\frac{\binom{3}{1}\binom{2}{1}}{\binom{5}{2}} = \frac{6}{10}$.
  • Urn $U_2$ now contains $2$ white balls and $1$ red ball.
  • Probability of drawing a white ball from $U_2$: $\frac{2}{3}$.

• Subcase 2.3: Two red balls are transferred.

  • Probability: $\frac{\binom{2}{2}}{\binom{5}{2}} = \frac{1}{10}$.
  • Urn $U_2$ now contains $1$ white ball and $2$ red balls.
  • Probability of drawing a white ball from $U_2$: $\frac{1}{3}$.

Thus, the conditional probability $P(W | T)$ is:
$P(W | T) = \left(\frac{3}{10} \times 1\right) + \left(\frac{6}{10} \times \frac{2}{3}\right) + \left(\frac{1}{10} \times \frac{1}{3}\right) = \frac{3}{10} + \frac{4}{10} + \frac{1}{30} = \frac{7}{10} + \frac{1}{30} = \frac{22}{30}$

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Total Probability $P(W)$

By applying the law of total probability:
$P(W) = P(H) \cdot P(W | H) + P(T) \cdot P(W | T)$

$P(W) = \left(\frac{1}{2} \times \frac{4}{5}\right) + \left(\frac{1}{2} \times \frac{22}{30}\right)$

$P(W) = \frac{1}{2} \times \left(\frac{24}{30} + \frac{22}{30}\right) = \frac{1}{2} \times \frac{46}{30} = \frac{23}{30}$

Hence, the correct option is B.