Stepwise
Start practisingPractise

WAEC 2014 · Paper 2 · Q5

A building contractor tendered for two independent contracts, XX and YY. The probability that he will win contract XX is 0.50.5 and that he will not win contract YY is 0.30.3. What is the probability that he will win:

  1. (a)

    both contracts;

  2. (b)

    exactly one of the contracts;

  3. (c)

    neither of the contracts?

Worked solution (try it first)
  1. Write all four probabilities first.
  2. P(wins X)=0.5P(\text{wins } X) = 0.5, so P(loses X)=0.5P(\text{loses } X) = 0.5.
  3. P(loses Y)=0.3P(\text{loses } Y) = 0.3, so P(wins Y)=1−0.3=0.7P(\text{wins } Y) = 1 - 0.3 = 0.7.
  4. The contracts are independent, so for "and" we multiply.

(a)

  1. Wins XX and wins YY: 0.5×0.7=0.350.5 \times 0.7 = 0.35.

(b)

  1. Exactly one happens in two ways: wins XX but not YY, 0.5×0.3=0.150.5 \times 0.3 = 0.15.
  2. Or wins YY but not XX, 0.5×0.7=0.350.5 \times 0.7 = 0.35.
  3. Add them: 0.15+0.35=0.50.15 + 0.35 = 0.5.

(c)

  1. Loses both: 0.5×0.3=0.150.5 \times 0.3 = 0.15.
  2. Check: both, exactly one and neither cover everything, and 0.35+0.5+0.15=10.35 + 0.5 + 0.15 = 1.

Report a problem with this question