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WAEC 2011 · Paper 2 · Q15

  1. (a)

    The probability that a patient recovers from a disease is 0.25. If 6 people are known to have contracted this disease, calculate the probability that: (i) more than three people survived; (ii) at most 2 people survived.

    Separate values with commas, e.g. 3, −2

  2. (b)

    Two distinct numbers are selected at random from the set P={2,3,4,5,6}P = \{2, 3, 4, 5, 6\}. Find the probability that: (i) the sum of the two numbers is 8; (ii) one of the numbers is a factor of the other.

    Separate values with commas, e.g. 3, −2

Worked solution (try it first)

(a)

  1. X∼B(6,0.25)X \sim B(6, 0.25), with q=0.75q = 0.75.

(i)

  1. More than three: P(4)+P(5)+P(6)=0.032959+0.004395+0.000244P(4) + P(5) + P(6) = 0.032959 + 0.004395 + 0.000244
    =0.0376= 0.0376.

(ii)

  1. At most 2: P(0)+P(1)+P(2)=0.177979+0.355957+0.296631P(0) + P(1) + P(2) = 0.177979 + 0.355957 + 0.296631
    =0.8306= 0.8306.

(b)

  1. There are (52)=10\binom52 = 10 pairs of distinct numbers.

(i)

  1. A sum of 8: {2,6}\{2, 6\} and {3,5}\{3, 5\}, so P=210=15P = \frac{2}{10} = \frac15.

(ii)

  1. One divides the other: {2,4}\{2, 4\}, {2,6}\{2, 6\} and {3,6}\{3, 6\}, so P=310P = \frac{3}{10}.

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