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WAEC 2014 · Paper 2 · Q13

The probabilities that Kofi, Kwasi and Ama will pass a certain examination are 910\frac{9}{10}, 45\frac45 and xx respectively. If the probability that only one of them will pass the examination is 950\frac{9}{50}, find the:

  1. (a)

    value of xx;

  2. (b)

    probability that at least one of them will pass the examination.

Worked solution (try it first)

(a)

  1. Only one passes means Kofi only, Kwasi only, or Ama only.
  2. In each case the other two fail.
  3. So 910×15(1−x)+110×45(1−x)+110×15x=950\frac{9}{10} \times \frac15(1 - x) + \frac{1}{10} \times \frac45(1 - x) + \frac{1}{10} \times \frac15x = \frac{9}{50}.
  4. Simplify: 950(1−x)+450(1−x)+150x=950\frac{9}{50}(1 - x) + \frac{4}{50}(1 - x) + \frac{1}{50}x = \frac{9}{50}.
  5. Multiply by 50: 13(1−x)+x=913(1 - x) + x = 9, so 13−12x=913 - 12x = 9.
  6. So 12x=412x = 4 and x=13x = \frac13.

(b)

  1. P(none pass) =110×15×23= \frac{1}{10} \times \frac15 \times \frac23
    =175= \frac{1}{75}.
  2. P(at least one passes) =1−175= 1 - \frac{1}{75}
    =7475= \frac{74}{75}
    ≈0.987\approx 0.987.

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