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WAEC 2024 · Paper 2 · Q5

A basket contains 3 gold-plated marbles, 4 diamond marbles and some silver marbles, all of the same size and shape. Two marbles were drawn from the basket at random one after the other without replacement. If the probability that both marbles were silver is 115\frac{1}{15}, find the number of silver marbles.

Worked solution (try it first)
  1. Let there be ss silver marbles, so s+7s + 7 marbles in all.
  2. Without replacement, the first is silver with probability ss+7\frac{s}{s + 7}, and the second with probability s−1s+6\frac{s - 1}{s + 6} (one silver marble and one marble in total have gone).
  3. So ss+7×s−1s+6=115\frac{s}{s + 7} \times \frac{s - 1}{s + 6} = \frac{1}{15}.
  4. Cross-multiply: 15s(s−1)=(s+7)(s+6)15s(s - 1) = (s + 7)(s + 6), so 15s2−15s=s2+13s+4215s^2 - 15s = s^2 + 13s + 42.
  5. Collect terms: 14s2−28s−42=014s^2 - 28s - 42 = 0.
  6. Divide by 14: s2−2s−3=0s^2 - 2s - 3 = 0, which factorises as (s−3)(s+1)=0(s - 3)(s + 1) = 0.
  7. So s=3s = 3 or s=−1s = -1.
  8. A number of marbles can't be negative, so there are 3 silver marbles.
  9. Check: 310×29=690\frac{3}{10} \times \frac{2}{9} = \frac{6}{90}
    =115= \frac{1}{15}.

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