A box contains 5 white and 3 black balls. Two balls are drawn at random, one after the other, without replacement. The probability that both the balls drawn are black is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{3}{28}\)
The box has 8 balls. For the first draw, 3 of the 8 balls are black:
\( \displaystyle P(\text{first black}) = \frac{3}{8} \)
The first ball is not replaced, so 7 balls are left, of which 2 are black:
\( \displaystyle P(\text{second black} \mid \text{first black}) = \frac{2}{7} \)
By the multiplication theorem:
\( \displaystyle P(\text{both black}) = \frac{3}{8} \times \frac{2}{7} \)
\( \displaystyle = \frac{6}{56} = \frac{3}{28} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{9}{64}\) — Incorrect. This is \(\frac{3}{8} \times \frac{3}{8}\), the answer with replacement.
Option (2) \(\displaystyle\frac{15}{56}\) — Incorrect. This is \(\frac{5}{8} \times \frac{3}{7}\), the probability of a white ball first and then a black ball.
Option (3) \(\displaystyle\frac{3}{8}\) — Incorrect. This is the probability that only the first ball is black.