Step-by-step Solution

Let's break down the problem to find the total probability of drawing a white ball from urn $U_2$.

1. Initial Setup

• Urn $U_1$ contains: 3 White (W) and 2 Red (R) balls. Total balls = 5.
• Urn $U_2$ contains: 1 White (W) ball. Total balls = 1.
• A fair coin is tossed, so the probability of getting a Head (H) is $P(H) = {1 \over 2}$ and the probability of getting a Tail (T) is $P(T) = {1 \over 2}$.

Let $E$ be the event that the ball drawn from $U_2$ is white. We need to find $P(E)$. We can use the Law of Total Probability: $P(E) = P(E|H) \cdot P(H) + P(E|T) \cdot P(T)$

2. Case 1: Head Appears (Event H) If a head appears, 1 ball is drawn from $U_1$ and put into $U_2$. We need to calculate $P(E|H)$, the probability of drawing a white ball from $U_2$ given that a head appeared.

• Subcase 2.1: A white ball is transferred from $U_1$ to $U_2$.

  • The probability of drawing a white ball from $U_1$ is $P(W_1) = {3 \over 5}$.
  • After transfer, $U_2$ will contain $1+1=2$ white balls. The total number of balls in $U_2$ is 2.
  • The probability of drawing a white ball from $U_2$ is now ${2 \over 2} = 1$.

• Subcase 2.2: A red ball is transferred from $U_1$ to $U_2$.

  • The probability of drawing a red ball from $U_1$ is $P(R_1) = {2 \over 5}$.
  • After transfer, $U_2$ will contain 1 white ball and 1 red ball. The total number of balls in $U_2$ is 2.
  • The probability of drawing a white ball from $U_2$ is now ${1 \over 2}$.

• Calculating $P(E|H)$: Using the law of total probability for this conditional event: $P(E|H) = P(\text{draw W from } U_2 | W_1 \text{ transferred}) \cdot P(W_1 \text{ transferred}) + P(\text{draw W from } U_2 | R_1 \text{ transferred}) \cdot P(R_1 \text{ transferred})$ $P(E|H) = (1) \cdot \left({3 \over 5}\right) + \left({1 \over 2}\right) \cdot \left({2 \over 5}\right) = {3 \over 5} + {1 \over 5} = {4 \over 5}$

3. Case 2: Tail Appears (Event T) If a tail appears, 2 balls are drawn from $U_1$ and put into $U_2$. We need to calculate $P(E|T)$, the probability of drawing a white ball from $U_2$ given that a tail appeared.

The total number of ways to draw 2 balls from $U_1$ (3W, 2R) is ${}^5C_2 = {5 \times 4 \over 2} = 10$.

• Subcase 3.1: Two white balls are transferred from $U_1$.

  • Number of ways to draw 2W from 3W is ${}^3C_2 = 3$. Probability = ${3 \over 10}$.
  • After transfer, $U_2$ contains $1+2=3$ white balls. Total balls = 3.
  • Probability of drawing a white ball from $U_2$ is ${3 \over 3} = 1$.

• Subcase 3.2: One white and one red ball are transferred from $U_1$.

  • Number of ways to draw 1W from 3W and 1R from 2R is ${}^3C_1 \times {}^2C_1 = 3 \times 2 = 6$. Probability = ${6 \over 10}$.
  • After transfer, $U_2$ contains $1+1=2$ white balls and 1 red ball. Total balls = 3.
  • Probability of drawing a white ball from $U_2$ is ${2 \over 3}$.

• Subcase 3.3: Two red balls are transferred from $U_1$.

  • Number of ways to draw 2R from 2R is ${}^2C_2 = 1$. Probability = ${1 \over 10}$.
  • After transfer, $U_2$ contains 1 white ball and 2 red balls. Total balls = 3.
  • Probability of drawing a white ball from $U_2$ is ${1 \over 3}$.

• Calculating $P(E|T)$: $P(E|T) = (1) \cdot \left({3 \over 10}\right) + \left({2 \over 3}\right) \cdot \left({6 \over 10}\right) + \left({1 \over 3}\right) \cdot \left({1 \over 10}\right)$ $P(E|T) = {3 \over 10} + {12 \over 30} + {1 \over 30} = {9 \over 30} + {12 \over 30} + {1 \over 30} = {22 \over 30}$

4. Final Calculation for P(E) Now, we substitute the values of $P(E|H)$ and $P(E|T)$ into the total probability formula: $P(E) = P(E|H) \cdot P(H) + P(E|T) \cdot P(T)$ $P(E) = \left({4 \over 5}\right) \cdot \left({1 \over 2}\right) + \left({22 \over 30}\right) \cdot \left({1 \over 2}\right)$ $P(E) = {4 \over 10} + {22 \over 60}$ To add these fractions, we find a common denominator, which is 60. $P(E) = {4 \times 6 \over 10 \times 6} + {22 \over 60} = {24 \over 60} + {22 \over 60} = {46 \over 60}$ Simplifying the fraction: $P(E) = {23 \over 30}$

The probability of the drawn ball from $U_2$ being white is ${{23} \over {30}}$. This matches option B.

Conclusion

The correct option is B.