Paper 4 (9164, Section A) · June 2014 · Probability

A bag contains 24 counters of which 6 are red, 8 are green and 10 are yellow. Three counters are taken from the bag at random without replacement.

Which calculation gives the probability that exactly 2 of the three counters taken are green?

\dfrac{\binom{8}{2}\binom{16}{1}}{\binom{24}{2}}

, choosing 2 from the 8 green ones and 1 from the 16 that are not green, over pairs drawn from 24

\dfrac{\binom{8}{2}}{\binom{24}{3}}

, choosing 2 counters from the 8 green ones and leaving the third counter out of the count altogether

\dfrac{\binom{8}{2}\binom{22}{1}}{\binom{24}{3}}

, choosing 2 from the 8 green ones and 1 from the 22 counters that remain once two have gone

\dfrac{\binom{8}{2}\binom{16}{1}}{\binom{24}{3}}

, choosing 2 from the 8 green ones and 1 from the 16 counters that are not green

Explanation

Exactly 2 green means two of the counters come from the 8 green ones and the third from the 16 that are not green, out of all the ways of taking 3 counters from 24.

$P = \dfrac{\binom{8}{2}\binom{16}{1}}{\binom{24}{3}} = \dfrac{28 \times 16}{2024} = \dfrac{448}{2024} = \dfrac{56}{253}$ .

The third counter must be accounted for, which rules out the first option. It must come from the 16 non-green counters, not from all 22 that are left, or the count includes samples with three green counters. And the denominator counts samples of 3, not of 2.

Derived from ZIMSEC Statistics Paper 4, June 2014, Q1

A bag contains 24 counters of which 6 are red, 8 are green and 10 are yellow. Three counters are taken from the bag at random without replacement.

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