A bag contains 8 balls,whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

Options

A. `2/5`
B. `2/7`
C. `1/7`
D. `1/5`

Answer

To solve the problem, we need to find the probability that the bag contains an equal number of white and black balls given that 2 white and 2 black balls were drawn from the bag.

Let's denote the events:
- Let \( A_1 \) be the event that the bag contains 4 white and 4 black balls.
- Let \( A_2 \) be the event that the bag contains 3 white and 5 black balls.
- Let \( A_3 \) be the event that the bag contains 2 white and 6 black balls.
- Let \( A_4 \) be the event that the bag contains 1 white and 7 black balls.
- Let \( A_5 \) be the event that the bag contains 0 white and 8 black balls.

We are interested in finding \( P(A_1 | E) \), where \( E \) is the event that 2 white and 2 black balls were drawn.

Using Bayes' theorem, we have:

\[ P(A_1 | E) = \frac{P(E | A_1) P(A_1)}{P(E)} \]

### Step 1: Calculate \( P(E | A_i) \) for each \( A_i \)

1. **For \( A_1 \) (4 white, 4 black)**:
\[ P(E | A_1) = \frac{\binom{4}{2} \binom{4}{2}}{\binom{8}{4}} = \frac{6 \cdot 6}{70} = \frac{36}{70} \]

2. **For \( A_2 \) (3 white, 5 black)**:
\[ P(E | A_2) = \frac{\binom{3}{2} \binom{5}{2}}{\binom{8}{4}} = \frac{3 \cdot 10}{70} = \frac{30}{70} \]

3. **For \( A_3 \) (2 white, 6 black)**:
\[ P(E | A_3) = \frac{\binom{2}{2} \binom{6}{2}}{\binom{8}{4}} = \frac{1 \cdot 15}{70} = \frac{15}{70} \]

4. **For \( A_4 \) (1 white, 7 black)**:
\[ P(E | A_4) = \frac{\binom{1}{2} \binom{7}{2}}{\binom{8}{4}} = 0 \quad (\text{impossible to draw 2 white}) \]

5. **For \( A_5 \) (0 white, 8 black)**:
\[ P(E | A_5) = \frac{\binom{0}{2} \binom{8}{2}}{\binom{8}{4}} = 0 \quad (\text{impossible to draw 2 white}) \]

### Step 2: Calculate \( P(A_i) \)

Assuming each configuration is equally likely, we have:
\[ P(A_1) = P(A_2) = P(A_3) = P(A_4) = P(A_5) = \frac{1}{5} \]

### Step 3: Calculate \( P(E) \)

Using the law of total probability:
\[ P(E) = P(E | A_1) P(A_1) + P(E | A_2) P(A_2) + P(E | A_3) P(A_3) + P(E | A_4) P(A_4) + P(E | A_5) P(A_5) \]
\[ = \frac{36}{70} \cdot \frac{1}{5} + \frac{30}{70} \cdot \frac{1}{5} + \frac{15}{70} \cdot \frac{1}{5} + 0 + 0 \]
\[ = \frac{36 + 30 + 15}{350} = \frac{81}{350} \]

### Step 4: Calculate \( P(A_1 | E) \)

Now substituting back into Bayes' theorem:
\[ P(A_1 | E) = \frac{P(E | A_1) P(A_1)}{P(E)} = \frac{\frac{36}{70} \cdot \frac{1}{5}}{\frac{81}{350}} = \frac{36 \cdot 350}{70 \cdot 5 \cdot 81} \]

Calculating this gives:
\[ = \frac{1260}{2835} = \frac{2}{7} \]

### Final Answer:
The probability that the bag contains an equal number of white and black balls is \( \frac{2}{7} \).

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