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 head appears then `1` ball is drawn at random from `U_1`, and put into `U_2,` . However, if 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,` .61 . The probability of the drawn ball from `U_2,` being white is

Options

A. `13/30`
B. `23/30`
C. `19/30`
D. `11/30`

Explanation

To solve the problem step by step, we will calculate the probability of drawing a white ball from urn \( U_2 \) after transferring balls from urn \( U_1 \) based on the outcome of the coin toss.

### Step 1: Identify the total number of balls in urns
- Urn \( U_1 \) contains 3 white and 2 red balls, totaling 5 balls.
- Urn \( U_2 \) initially contains 1 white ball.

### Step 2: Determine the outcomes based on the coin toss
- If the coin shows heads (H), 1 ball is drawn from \( U_1 \) and placed in \( U_2 \).
- If the coin shows tails (T), 2 balls are drawn from \( U_1 \) and placed in \( U_2 \).

### Step 3: Calculate the probabilities for each case
1. **Case 1: Coin shows Heads (H)**
- Probability of getting heads: \( P(H) = \frac{1}{2} \)
- The probability of drawing a white ball from \( U_1 \): \( P(W|H) = \frac{3}{5} \)
- If a white ball is drawn, \( U_2 \) will have 2 white balls (1 original + 1 drawn).
- The probability of drawing a white ball from \( U_2 \) in this case:
\[ P(W|H) = \frac{2}{2} = 1 \]
- The probability of drawing a red ball from \( U_1 \): \( P(R|H) = \frac{2}{5} \)
- If a red ball is drawn, \( U_2 \) will have 1 white and 1 red ball.
- The probability of drawing a white ball from \( U_2 \) in this case:
\[ P(W|H) = \frac{1}{2} \]

2. **Case 2: Coin shows Tails (T)**
- Probability of getting tails: \( P(T) = \frac{1}{2} \)
- The probability of drawing 2 balls from \( U_1 \):
- The combinations of drawing 2 white balls: \( \frac{3C2}{5C2} = \frac{3}{10} \)
- The combinations of drawing 1 white and 1 red ball: \( \frac{3C1 \cdot 2C1}{5C2} = \frac{6}{10} = \frac{3}{5} \)
- The combinations of drawing 2 red balls: \( \frac{2C2}{5C2} = \frac{1}{10} \)

- If 2 white balls are drawn, \( U_2 \) will have 3 white balls:
\[ P(W|T, 2W) = \frac{3}{3} = 1 \]
- If 1 white and 1 red ball are drawn, \( U_2 \) will have 2 white and 1 red ball:
\[ P(W|T, 1W, 1R) = \frac{2}{3} \]
- If 2 red balls are drawn, \( U_2 \) will have 1 white and 2 red balls:
\[ P(W|T, 2R) = \frac{1}{3} \]

### Step 4: Combine the probabilities using the law of total probability
Now, we can combine these probabilities:
\[ P(W) = P(W|H)P(H) + P(W|T)P(T) \]
Where:
- \( P(W|H) = P(W|H, W)P(W|H) + P(W|H, R)P(R|H) \)
- \( P(W|T) = P(W|T, 2W)P(2W|T) + P(W|T, 1W, 1R)P(1W, 1R|T) + P(W|T, 2R)P(2R|T) \)

Calculating each part:
- For heads:
\[ P(W|H) = 1 \cdot \frac{3}{5} + \frac{1}{2} \cdot \frac{2}{5} = \frac{3}{5} + \frac{1}{5} = \frac{4}{5} \]
- For tails:
\[ P(W|T) = 1 \cdot \frac{3}{10} + \frac{2}{3} \cdot \frac{3}{5} + \frac{1}{3} \cdot \frac{1}{10} = \frac{3}{10} + \frac{6}{15} + \frac{1}{30} = \frac{9}{30} + \frac{12}{30} + \frac{1}{30} = \frac{22}{30} \]

### Final Probability Calculation
Now substituting back into the total probability:
\[ P(W) = \frac{1}{2} \cdot \frac{4}{5} + \frac{1}{2} \cdot \frac{22}{30} \]
Calculating this gives:
\[ P(W) = \frac{2}{5} + \frac{11}{30} = \frac{12}{30} + \frac{11}{30} = \frac{23}{30} \]

### Conclusion
The probability of drawing a white ball from \( U_2 \) is \( \frac{23}{30} \).