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, we will break it down into steps and calculate the required probability that a ball drawn from urn \( U_2 \) is white.

### Step 1: Understand the setup
- **Urn \( U_1 \)** contains 3 white balls and 2 red balls.
- **Urn \( U_2 \)** initially contains 1 white ball.
- A fair coin is tossed. If it lands heads, 1 ball is drawn from \( U_1 \) and placed in \( U_2 \). If it lands tails, 2 balls are drawn from \( U_1 \) and placed in \( U_2 \).

### Step 2: Calculate probabilities for each scenario based on the coin toss

#### Case 1: Coin shows Heads
- Probability of heads = \( \frac{1}{2} \)
- 1 ball is drawn from \( U_1 \):
- Probability of drawing a white ball = \( \frac{3}{5} \)
- Probability of drawing a red ball = \( \frac{2}{5} \)

**After the draw:**
- If a white ball is drawn: \( U_2 \) will have 2 white balls (1 original + 1 drawn).
- If a red ball is drawn: \( U_2 \) will have 1 white ball and 1 red ball.

#### Case 2: Coin shows Tails
- Probability of tails = \( \frac{1}{2} \)
- 2 balls are drawn from \( U_1 \):
- Possible combinations:
1. 2 white balls
2. 1 white and 1 red ball
3. 2 red balls

**Calculating probabilities for each combination:**
1. **2 white balls:**
- Probability = \( \frac{3}{5} \times \frac{2}{4} = \frac{3}{10} \)
- \( U_2 \) will have 3 white balls.

2. **1 white and 1 red ball:**
- Probability = \( \frac{3}{5} \times \frac{2}{4} + \frac{2}{5} \times \frac{3}{4} = \frac{3}{10} + \frac{3}{10} = \frac{3}{5} \)
- \( U_2 \) will have 2 white balls and 1 red ball.

3. **2 red balls:**
- Probability = \( \frac{2}{5} \times \frac{1}{4} = \frac{1}{10} \)
- \( U_2 \) will have 1 white ball and 2 red balls.

### Step 3: Combine probabilities for \( U_2 \)
Now we compute the total probability of drawing a white ball from \( U_2 \) for both cases.

#### Case 1: Coin shows Heads
- Probability of drawing a white ball from \( U_2 \):
- If a white ball is drawn: \( P(W|W) = 1 \)
- If a red ball is drawn: \( P(W|R) = \frac{1}{2} \)

Total probability from heads:
\[ P(W|H) = \frac{1}{2} \left( \frac{3}{5} \cdot 1 + \frac{2}{5} \cdot \frac{1}{2} \right) = \frac{1}{2} \left( \frac{3}{5} + \frac{1}{5} \right) = \frac{1}{2} \cdot \frac{4}{5} = \frac{2}{5} \]

#### Case 2: Coin shows Tails
- Probability of drawing a white ball from \( U_2 \):
1. For 2 white balls: \( P(W|2W) = 1 \)
2. For 1 white and 1 red: \( P(W|1W, 1R) = \frac{2}{3} \)
3. For 2 red balls: \( P(W|2R) = 0 \)

Total probability from tails:
\[ P(W|T) = \frac{1}{2} \left( \frac{3}{10} \cdot 1 + \frac{3}{5} \cdot \frac{2}{3} + \frac{1}{10} \cdot 0 \right) = \frac{1}{2} \left( \frac{3}{10} + \frac{6}{10} \right) = \frac{1}{2} \cdot \frac{9}{10} = \frac{9}{20} \]

### Step 4: Total probability of drawing a white ball from \( U_2 \)
Combine the probabilities from both cases:
\[ P(W) = P(W|H) + P(W|T) = \frac{2}{5} \cdot \frac{1}{2} + \frac{9}{20} \cdot \frac{1}{2} \]
\[ P(W) = \frac{2}{10} + \frac{9}{40} = \frac{8}{40} + \frac{9}{40} = \frac{17}{40} \]

### Final Answer
The probability of the drawn ball from \( U_2 \) being white is \( \frac{17}{40} \).