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WAEC 2009 · Paper 2 · Q6

The chances of three hunters hitting a target are 13\frac13, 23\frac23 and 15\frac15 respectively. If they fire independently at the target, find the probability that

  1. (a)

    at least one of them hits the target;

  2. (b)

    only two of them hit the target.

Worked solution (try it first)

(a)

  1. The probabilities of missing are 1−13=231 - \frac13 = \frac23, 1−23=131 - \frac23 = \frac13 and 1−15=451 - \frac15 = \frac45.
  2. The hunters fire independently, so P(none hits)=23×13×45P(\text{none hits}) = \frac23 \times \frac13 \times \frac45
    =845= \frac{8}{45}.
  3. P(at least one)=1−845P(\text{at least one}) = 1 - \frac{8}{45}
    =3745= \frac{37}{45}.

(b)

  1. First two hit, third misses: 13×23×45=845\frac13 \times \frac23 \times \frac45 = \frac{8}{45}.
  2. First and third hit, second misses: 13×13×15=145\frac13 \times \frac13 \times \frac15 = \frac{1}{45}.
  3. Second and third hit, first misses: 23×23×15=445\frac23 \times \frac23 \times \frac15 = \frac{4}{45}.
  4. Add the three cases: 8+1+445=1345\frac{8 + 1 + 4}{45} = \frac{13}{45}.

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