A box contains 1 black and 1 white ball. A ball is drawn randomly and replaced in the box with an additional ball of the same colour, then a second ball is drawn randomly from the box containing 3 balls. The probability that the first drawn ball was white given that at least one of the two balls drawn was white is

Options

A. `(1)/(2)`
B. `(3)/(4)`
C. `(4)/(5)`
D. `(5)/(11)`

Answer

To solve the problem, we need to find the probability that the first drawn ball was white given that at least one of the two balls drawn was white. We will use the concept of conditional probability.

### Step-by-Step Solution:

1. **Identify the Initial Setup**:
- The box contains 1 black ball (B) and 1 white ball (W).
- Total balls initially = 2 (1W + 1B).

2. **Draw the First Ball**:
- The probability of drawing a white ball (W) first:
\[ P(W) = \frac{1}{2} \]
- The probability of drawing a black ball (B) first:
\[ P(B) = \frac{1}{2} \]

3. **Replace and Add a Ball**:
- If the first ball drawn is white (W):
- The box now contains 2 white balls and 1 black ball (2W + 1B).
- If the first ball drawn is black (B):
- The box now contains 1 white ball and 2 black balls (1W + 2B).

4. **Draw the Second Ball**:
- **Case 1**: First ball was white (W):
- Probability of drawing a white ball (W) second:
\[ P(W \text{ second} | W \text{ first}) = \frac{2}{3} \]
- Probability of drawing a black ball (B) second:
\[ P(B \text{ second} | W \text{ first}) = \frac{1}{3} \]

- **Case 2**: First ball was black (B):
- Probability of drawing a white ball (W) second:
\[ P(W \text{ second} | B \text{ first}) = \frac{1}{3} \]
- Probability of drawing a black ball (B) second:
\[ P(B \text{ second} | B \text{ first}) = \frac{2}{3} \]

5. **Calculate the Probability of At Least One White Ball**:
- We need to find the total probability of getting at least one white ball in the two draws. This can happen in the following ways:
- First W, second W: \( P(W \text{ first}) \times P(W \text{ second} | W \text{ first}) = \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \)
- First W, second B: \( P(W \text{ first}) \times P(B \text{ second} | W \text{ first}) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \)
- First B, second W: \( P(B \text{ first}) \times P(W \text{ second} | B \text{ first}) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \)

- Total probability of at least one white ball:
\[ P(\text{at least one W}) = P(W, W) + P(W, B) + P(B, W) = \frac{1}{3} + \frac{1}{6} + \frac{1}{6} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]

6. **Calculate the Conditional Probability**:
- We want to find \( P(W \text{ first} | \text{at least one W}) \):
\[ P(W \text{ first} | \text{at least one W}) = \frac{P(W \text{ first and at least one W})}{P(\text{at least one W})} \]
- The numerator \( P(W \text{ first and at least one W}) = P(W \text{ first}) \) since if the first ball is white, at least one white ball is guaranteed:
\[ P(W \text{ first}) = \frac{1}{2} \]
- Therefore,
\[ P(W \text{ first} | \text{at least one W}) = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4} \]

### Final Answer:
The probability that the first drawn ball was white given that at least one of the two balls drawn was white is:
\[ \frac{3}{4} \]