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:
\[ P(4W4B / 2W2B) = \frac{P(4W4B) \times P(2W2B / 4W4B)}{P(4W4B) \times P(2W2B / 4W4B) + P(3W5B) \times P(2W2B / 3W5B) + \dots + P(0W8B) \times P(2W2B / 0W8B)} \] \[ = \frac{\frac{1}{5} \times \binom{4}{2} \times \binom{4}{2} / \binom{8}{4}}{\frac{1}{5} \times \binom{2}{2} \times \binom{6}{2} / \binom{8}{4} + \frac{1}{5} \times \binom{3}{2} \times \binom{5}{2} / \binom{8}{4} + \dots + \frac{1}{5} \times \binom{6}{2} \times \binom{2}{2} / \binom{8}{4}} \] \[ = \frac{\frac{1}{5} \times \frac{4C_2 \times 4C_2}{8C_4}}{\frac{1}{5} \times \frac{2C_2 \times 6C_2}{8C_4} + \frac{1}{5} \times \frac{3C_2 \times 5C_2}{8C_4} + \dots + \frac{1}{5} \times \frac{6C_2 \times 2C_2}{8C_4}} \] \[ = \frac{\frac{1}{5} \times \frac{6 \times 6}{70}}{\frac{1}{5} \times \frac{15}{70} + \frac{1}{5} \times \frac{30}{70} + \dots + \frac{1}{5} \times \frac{15}{70}} \] \[ = \frac{\frac{1}{5} \times \frac{6 \times 6}{70}}{\frac{1}{5} \times \frac{15}{70} + \frac{1}{5} \times \frac{30}{70} + \dots + \frac{1}{5} \times \frac{15}{70}} \] \[ = \frac{\frac{6}{70}}{\frac{15}{70} + \frac{30}{70} + \frac{30}{70} + \frac{15}{70}} \] \[ = \frac{\frac{6}{70}}{\frac{90}{70}} = \frac{6}{90} = \frac{2}{7}. \]