Let U1 and U2 be two urns such that U1 contains 3 white and 2 red balls, and U2 contains only 1 white ball. A fair coin is tossed. If head appears then 1 ball is drawn at random from U1 and put into U2. However, if tail appears then 2 balls are drawn at random from U1 and put into U2. Now 1 ball is drawn at random from U2. Given that the drawn ball from U2 is white, the probability that head appeared on the coin is

Options

A. `17/23`
B. `11/23`
C. `15/23`
D. `12/23`

Explanation

To solve the problem step by step, we will use the concepts of conditional probability and the law of total probability.

### Step 1: Define Events
Let:
- \( H \): Event that the coin shows heads.
- \( T \): Event that the coin shows tails.
- \( W \): Event that a white ball is drawn from urn \( U_2 \).

### Step 2: Calculate Probabilities for Each Case
1. **Case when the coin shows heads (H)**:
- If heads appears, we draw 1 ball from \( U_1 \) (which has 3 white and 2 red balls) and put it into \( U_2 \) (which initially has 1 white ball).
- The probability of drawing a white ball from \( U_1 \) is \( \frac{3}{5} \) (3 white out of 5 total).
- If we draw a white ball, \( U_2 \) will have 2 white balls. The probability of drawing a white ball from \( U_2 \) is now \( \frac{2}{2} = 1 \).
- If we draw a red ball, \( U_2 \) will have 1 white and 1 red ball. The probability of drawing a white ball from \( U_2 \) is \( \frac{1}{2} \).

Therefore, the total probability of drawing a white ball from \( U_2 \) given heads is:
\[ P(W | H) = P(W | H, \text{white from } U_1) \cdot P(\text{white from } U_1) + P(W | H, \text{red from } U_1) \cdot P(\text{red from } U_1) \]
\[ P(W | H) = 1 \cdot \frac{3}{5} + \frac{1}{2} \cdot \frac{2}{5} = \frac{3}{5} + \frac{1}{5} = \frac{4}{5} \]

2. **Case when the coin shows tails (T)**:
- If tails appears, we draw 2 balls from \( U_1 \).
- The possible combinations of drawing 2 balls from \( U_1 \) can be:
- 2 white balls: Probability = \( \frac{3}{5} \cdot \frac{2}{4} = \frac{3}{10} \)
- 1 white and 1 red ball: Probability = \( \frac{3}{5} \cdot \frac{2}{4} + \frac{2}{5} \cdot \frac{3}{4} = \frac{3}{10} + \frac{3}{10} = \frac{3}{5} \)
- 2 red balls: Probability = \( \frac{2}{5} \cdot \frac{1}{4} = \frac{1}{10} \)

Now, the probabilities of drawing a white ball from \( U_2 \) in each case:
- If we draw 2 white balls, \( U_2 \) has 3 white balls: \( P(W | T, 2W) = 1 \)
- If we draw 1 white and 1 red, \( U_2 \) has 2 white and 1 red: \( P(W | T, 1W, 1R) = \frac{2}{3} \)
- If we draw 2 red balls, \( U_2 \) has 1 white and 2 red: \( P(W | T, 2R) = \frac{1}{3} \)

Therefore, the total probability of drawing a white ball from \( U_2 \) given tails is:
\[ P(W | T) = P(W | T, 2W) \cdot P(2W) + P(W | T, 1W, 1R) \cdot P(1W, 1R) + P(W | T, 2R) \cdot P(2R) \]
\[ P(W | T) = 1 \cdot \frac{3}{10} + \frac{2}{3} \cdot \frac{3}{5} + \frac{1}{3} \cdot \frac{1}{10} \]
\[ P(W | T) = \frac{3}{10} + \frac{2}{5} + \frac{1}{30} = \frac{3}{10} + \frac{12}{30} + \frac{1}{30} = \frac{16}{30} = \frac{8}{15} \]

### Step 3: Use the Law of Total Probability
Now, we can find \( P(W) \):
\[ P(W) = P(W | H) \cdot P(H) + P(W | T) \cdot P(T) \]
\[ P(W) = \frac{4}{5} \cdot \frac{1}{2} + \frac{8}{15} \cdot \frac{1}{2} = \frac{4}{10} + \frac{8}{30} \]
Finding a common denominator (30):
\[ P(W) = \frac{12}{30} + \frac{8}{30} = \frac{20}{30} = \frac{2}{3} \]

### Step 4: Calculate \( P(H | W) \)
Now we apply Bayes' theorem:
\[ P(H | W) = \frac{P(W | H) \cdot P(H)}{P(W)} \]
\[ P(H | W) = \frac{\frac{4}{5} \cdot \frac{1}{2}}{\frac{2}{3}} = \frac{\frac{4}{10}}{\frac{2}{3}} = \frac{4}{10} \cdot \frac{3}{2} = \frac{12}{20} = \frac{3}{5} \]

### Final Answer
Thus, the probability that heads appeared on the coin given that the drawn ball from \( U_2 \) is white is:
\[ \boxed{\frac{3}{5}} \]