WAEC 2019 · Paper 2 · Q5

A fair die with faces 1, 2, 3, 4, 5 and 6 is tossed twice. Calculate the probability that the sum of the numbers that show up is:

  1. (a)

    a multiple of 3;

  2. (b)

    between 3 and 6.

Worked solution (try it first)
  1. Two throws give 36 equally likely outcomes.

(a)

  1. Sums that are multiples of 3: 3 (2 ways), 6 (5 ways), 9 (4 ways) and 12 (1 way): 12 outcomes, so P=1236=13P = \frac{12}{36} = \frac13.

(b)

  1. Between 3 and 6 means a sum of 4 (3 ways) or 5 (4 ways): 7 outcomes, so P=736P = \frac{7}{36}.

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