A bag contains 4 red and 6 black balls. Two balls are drawn in succession without replacement. The probability that the first is red and the second is black is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{4}{15}\)
There are 10 balls, 4 of them red:
\( \displaystyle P(\text{first red}) = \frac{4}{10} \)
Without replacement, 9 balls remain after a red one is drawn, and all 6 black balls are still there:
\( \displaystyle P(\text{second black} \mid \text{first red}) = \frac{6}{9} \)
By the multiplication rule,
\( \displaystyle P = \frac{4}{10}\times\frac{6}{9} = \frac{24}{90} = \frac{4}{15} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{2}{5}\) — Incorrect. This is only \(P(\text{first red}) = \frac{4}{10}\).
Option (2) \(\displaystyle\frac{2}{3}\) — Incorrect. This is only the conditional probability \(\frac{6}{9}\) of black on the second draw.
Option (4) \(\displaystyle\frac{6}{25}\) — Incorrect. This is \(\frac{4}{10}\times\frac{6}{10}\), the probability with replacement.