WAEC 2018 · Paper 2 · Q12

Number of heads 0 1 2 3 4 5 6 7 8 9 10
Frequency 2 7 23 36 11 61 100 12 8 5 3

Ten coins were tossed together a number of times. The distribution of the number of heads obtained is given in the table. Calculate, correct to three decimal places, the:

  1. (a)

    mean number of heads;

  2. (b)

    probability of getting an even number of heads;

  3. (c)

    probability of getting an odd number of heads.

Worked solution (try it first)

(a)

  1. ∑f=268\sum f = 268 and ∑fx=1333\sum fx = 1333, so the mean is 1333268≈4.974\dfrac{1333}{268} \approx 4.974.

(b)

  1. Even numbers of heads are 0, 2, 4, 6, 8 and 10: 2+23+11+100+8+3=1472 + 23 + 11 + 100 + 8 + 3 = 147, so P=147268≈0.549P = \dfrac{147}{268} \approx 0.549.

(c)

  1. Odd: 7+36+61+12+5=1217 + 36 + 61 + 12 + 5 = 121, so P=121268≈0.451P = \dfrac{121}{268} \approx 0.451.

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