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WAEC 2012 · Paper 2 · Q7

Number of heads 0 1 2 3 4 5 6 7 8
Frequency 3 8 24 37 10 60 79 11 9

Eight coins were tossed together several times and the number of times heads appeared was recorded as shown. Find the probability of obtaining:

  1. (a)

    exactly 8 heads;

  2. (b)

    between 2 and 5 heads;

  3. (c)

    at most 1 head.

Worked solution (try it first)
  1. Add the frequencies: 3+8+24+37+10+60+79+11+9=2413 + 8 + 24 + 37 + 10 + 60 + 79 + 11 + 9 = 241 tosses.

(a)

  1. Exactly 8 heads: 9241≈0.0373\dfrac{9}{241} \approx 0.0373.

(b)

  1. Between 2 and 5 means 3 or 4 heads: 37+10241=47241\dfrac{37 + 10}{241} = \dfrac{47}{241}
    ≈0.1950\approx 0.1950.

(c)

  1. At most 1 head means 0 or 1: 3+8241=11241\dfrac{3 + 8}{241} = \dfrac{11}{241}
    ≈0.0456\approx 0.0456.

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