In the above example, given that the ball drawn from `U_(2)` 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, we will use Bayes' theorem, which allows us to find the conditional probability of an event given another event has occurred.

### Step-by-Step Solution:

1. **Define Events**:
- Let \( H \) be the event that the coin shows heads.
- Let \( W \) be the event that a white ball is drawn from \( U_2 \).

2. **Find Required Probability**:
- We need to find \( P(H | W) \), the probability that the coin shows heads given that a white ball is drawn.

3. **Apply Bayes' Theorem**:
- According to Bayes' theorem:
\[ P(H | W) = \frac{P(W | H) \cdot P(H)}{P(W)} \]

4. **Calculate Each Component**:
- **Calculate \( P(W | H) \)**: This is the probability of drawing a white ball from \( U_2 \) given that the coin shows heads. Let's assume \( U_2 \) contains 3 white balls and 2 black balls:
\[ P(W | H) = \frac{3}{5} \]
- **Calculate \( P(H) \)**: This is the prior probability of the coin showing heads. Assuming the coin is fair:
\[ P(H) = \frac{1}{2} \]
- **Calculate \( P(W) \)**: This is the total probability of drawing a white ball. It can be calculated using the law of total probability:
\[ P(W) = P(W | H) \cdot P(H) + P(W | T) \cdot P(T) \]
where \( T \) is the event that the coin shows tails. Assuming \( U_2 \) has 2 white balls and 3 black balls:
\[ P(W | T) = \frac{2}{5} \]
and
\[ P(T) = \frac{1}{2} \]
Thus,
\[ P(W) = \left(\frac{3}{5} \cdot \frac{1}{2}\right) + \left(\frac{2}{5} \cdot \frac{1}{2}\right) = \frac{3}{10} + \frac{2}{10} = \frac{5}{10} = \frac{1}{2} \]

5. **Substitute Values into Bayes' Theorem**:
- Now we can substitute the values into Bayes' theorem:
\[ P(H | W) = \frac{P(W | H) \cdot P(H)}{P(W)} = \frac{\left(\frac{3}{5}\right) \cdot \left(\frac{1}{2}\right)}{\frac{1}{2}} = \frac{3}{5} \]

6. **Final Answer**:
- Therefore, the probability that the coin shows heads given that a white ball is drawn is:
\[ P(H | W) = \frac{3}{5} \]