Question

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:

• A. \(\frac{2}{5}\)
• B. \(\frac{2}{7}\)
• C. \(\frac{1}{7}\)
• D. \(\frac{1}{5}\)

Answer 1

The Correct Option is B

Let us denote 4W4B as the case where the bag contains 4 white and 4 black balls. The probability of drawing 2 white and 2 black balls from such a bag is given by:
\[ P(4W4B / 2W2B) = \frac{P(4W4B) \times P(2W2B / 4W4B)}{P(4W4B) \times P(2W2B / 4W4B) + P(3W5B) \times P(2W2B / 3W5B) + \dots + P(0W8B) \times P(2W2B / 0W8B)} \] \[ = \frac{\frac{1}{5} \times \binom{4}{2} \times \binom{4}{2} / \binom{8}{4}}{\frac{1}{5} \times \binom{2}{2} \times \binom{6}{2} / \binom{8}{4} + \frac{1}{5} \times \binom{3}{2} \times \binom{5}{2} / \binom{8}{4} + \dots + \frac{1}{5} \times \binom{6}{2} \times \binom{2}{2} / \binom{8}{4}} \] \[ = \frac{\frac{1}{5} \times \frac{4C_2 \times 4C_2}{8C_4}}{\frac{1}{5} \times \frac{2C_2 \times 6C_2}{8C_4} + \frac{1}{5} \times \frac{3C_2 \times 5C_2}{8C_4} + \dots + \frac{1}{5} \times \frac{6C_2 \times 2C_2}{8C_4}} \] \[ = \frac{\frac{1}{5} \times \frac{6 \times 6}{70}}{\frac{1}{5} \times \frac{15}{70} + \frac{1}{5} \times \frac{30}{70} + \dots + \frac{1}{5} \times \frac{15}{70}} \] \[ = \frac{\frac{1}{5} \times \frac{6 \times 6}{70}}{\frac{1}{5} \times \frac{15}{70} + \frac{1}{5} \times \frac{30}{70} + \dots + \frac{1}{5} \times \frac{15}{70}} \] \[ = \frac{\frac{6}{70}}{\frac{15}{70} + \frac{30}{70} + \frac{30}{70} + \frac{15}{70}} \] \[ = \frac{\frac{6}{70}}{\frac{90}{70}} = \frac{6}{90} = \frac{2}{7}. \] 

Answer 2

Find the probability that the bag contains an equal number of white and black balls, given that 4 balls drawn at random without replacement consist of 2 white and 2 black balls.

Concept Used:

We apply Bayes' theorem. Let \( W \) be the number of white balls in the bag. The bag has 8 balls, so possible values for \( W \) are 0,1,2,...,8. However, since we drew 2 white and 2 black balls, \( W \) must be at least 2 and at most 6. We assume all compositions of colors are equally likely a priori.

Step-by-Step Solution:

Step 1: Define events:

• Let \( E \): Event that a sample of 4 balls has 2 white and 2 black balls.
• Let \( A \): Event that the bag contains equal white and black balls, i.e., \( W = 4 \).

We want \( P(A|E) \).

Step 2: Apply Bayes' theorem:

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

Assuming uniform prior over possible \( W \), \( P(A) = \frac{1}{9} \) (since \( W \) can be 0,...,8). But given we drew 2 white and 2 black, possible \( W \) values are 2,3,4,5,6. So prior over these 5 possibilities is uniform: \( P(W = w) = \frac{1}{5} \) for \( w = 2,3,4,5,6 \).

Step 3: Compute \( P(E|W = w) \):

Given \( W = w \), number of black balls = \( 8 - w \).

\[ P(E|W = w) = \frac{\binom{w}{2} \binom{8-w}{2}}{\binom{8}{4}} \]

Step 4: Compute \( P(E) \) using law of total probability:

\[ P(E) = \sum_{w=2}^{6} P(E|W=w) \cdot P(W=w) = \frac{1}{5} \cdot \frac{1}{\binom{8}{4}} \sum_{w=2}^{6} \binom{w}{2} \binom{8-w}{2} \]

Calculate the sum:

• \( w=2 \): \( \binom{2}{2} \binom{6}{2} = 1 \times 15 = 15 \)
• \( w=3 \): \( \binom{3}{2} \binom{5}{2} = 3 \times 10 = 30 \)
• \( w=4 \): \( \binom{4}{2} \binom{4}{2} = 6 \times 6 = 36 \)
• \( w=5 \): \( \binom{5}{2} \binom{3}{2} = 10 \times 3 = 30 \)
• \( w=6 \): \( \binom{6}{2} \binom{2}{2} = 15 \times 1 = 15 \)

Sum = \( 15 + 30 + 36 + 30 + 15 = 126 \).

Also, \( \binom{8}{4} = 70 \).

\[ P(E) = \frac{1}{5} \cdot \frac{126}{70} = \frac{126}{350} = \frac{63}{175} = \frac{9}{25} \]

Step 5: Compute \( P(E|A) \):

For \( W = 4 \):

\[ P(E|A) = \frac{\binom{4}{2} \binom{4}{2}}{\binom{8}{4}} = \frac{6 \times 6}{70} = \frac{36}{70} = \frac{18}{35} \]

Step 6: Apply Bayes' theorem:

\[ P(A|E) = \frac{P(E|A) \cdot P(A)}{P(E)} = \frac{\frac{18}{35} \cdot \frac{1}{5}}{\frac{9}{25}} = \frac{18}{35 \cdot 5} \cdot \frac{25}{9} = \frac{18 \cdot 25}{35 \cdot 5 \cdot 9} \]

Simplify:

\[ = \frac{450}{1575} = \frac{450 \div 225}{1575 \div 225} = \frac{2}{7} \]

Therefore, the required probability is \( \frac{2}{7} \).