A committee of 3 is chosen at random from a group of 5 boys and 3 girls. If it is known that the committee has at least one girl, then the probability that it has exactly one girl is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{15}{23}\)
All selections of 3 from the 8 persons are equally likely:
\({}^{8}C_{3} = 56\)
The committees with no girl are those of 3 boys only:
\({}^{5}C_{3} = 10\)
So the given condition leaves a reduced sample space of committees with at least one girl:
\(56-10 = 46\)
The committees with exactly one girl (1 girl and 2 boys) are
\({}^{3}C_{1}\times{}^{5}C_{2} = 3\times 10 = 30\)
All of these have at least one girl, so
\( \displaystyle P = \frac{30}{46} = \frac{15}{23} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{15}{28}\) — Incorrect. This is \(\frac{30}{56}\), the probability of exactly one girl without the given condition.
Option (2) \(\displaystyle\frac{23}{28}\) — Incorrect. This is \(\frac{46}{56}\), the probability of at least one girl, the condition itself.
Option (3) \(\displaystyle\frac{8}{23}\) — Incorrect. This is \(\frac{16}{46}\), the probability of two or more girls given at least one girl.