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 step by step, we will use the concept of conditional probability. 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.

### Step 1: Define Events
Let:
- Event \( B \): The first ball drawn is white.
- Event \( A \): At least one of the two balls drawn is white.

We need to find \( P(B|A) \), which is given by the formula:
\[ P(B|A) = \frac{P(B \cap A)}{P(A)} \]

### Step 2: Calculate \( P(A) \)
To find \( P(A) \), we need to consider the possible outcomes of drawing the balls.

1. **First ball drawn is white (W)**:
- The box will then contain 1 black and 2 white balls.
- The second ball can be either white or black:
- Probability of drawing white second: \( \frac{2}{3} \)
- Probability of drawing black second: \( \frac{1}{3} \)

Thus, the outcomes are:
- WW (White, White) with probability \( \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \)
- WB (White, Black) with probability \( \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \)

2. **First ball drawn is black (B)**:
- The box will then contain 2 black and 1 white ball.
- The second ball can be either white or black:
- Probability of drawing white second: \( \frac{1}{3} \)
- Probability of drawing black second: \( \frac{2}{3} \)

Thus, the outcomes are:
- BW (Black, White) with probability \( \frac{1}{2} \times \frac{1}{3} = \frac{1}{6} \)
- BB (Black, Black) with probability \( \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} \)

Now, we can summarize the probabilities:
- WW: \( \frac{1}{3} \)
- WB: \( \frac{1}{6} \)
- BW: \( \frac{1}{6} \)
- BB: \( \frac{1}{3} \)

Now, to find \( P(A) \):
- At least one white ball: WW, WB, BW
\[ P(A) = P(WW) + P(WB) + P(BW) = \frac{1}{3} + \frac{1}{6} + \frac{1}{6} = \frac{1}{3} + \frac{2}{6} = \frac{1}{3} + \frac{1}{3} = \frac{2}{3} \]

### Step 3: Calculate \( P(B \cap A) \)
For \( P(B \cap A) \), we consider the scenarios where the first ball drawn is white and at least one of the two balls drawn is white:
- The outcomes that satisfy both \( B \) and \( A \) are WW and WB.

Calculating the probabilities:
- WW: \( \frac{1}{3} \)
- WB: \( \frac{1}{6} \)

Thus:
\[ P(B \cap A) = P(WW) + P(WB) = \frac{1}{3} + \frac{1}{6} = \frac{2}{6} + \frac{1}{6} = \frac{3}{6} = \frac{1}{2} \]

### Step 4: Calculate \( P(B|A) \)
Now we can substitute back into the conditional probability formula:
\[ P(B|A) = \frac{P(B \cap A)}{P(A)} = \frac{\frac{1}{2}}{\frac{2}{3}} = \frac{1}{2} \times \frac{3}{2} = \frac{3}{4} \]

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