Paper 1 · conditional probability

A student travels to school by bus, car, or on foot with probabilities $\frac{1}{6}$ , $\frac{1}{3}$ , and $\frac{1}{2}$ respectively. The probability of being late is $\frac{1}{5}$ if by bus, $\frac{1}{4}$ if by car, and $\frac{1}{20}$ if on foot. What is the probability that the student is early (not late) for school?

\frac{17}{120}

\frac{103}{120}

\frac{1}{2}

\frac{19}{40}

Explanation: By the law of total probability, P(late) = (1/6)(1/5) + (1/3)(1/4) + (1/2)(1/20) = 4/120 + 10/120 + 3/120 = 17/120. So P(early) = 1 - 17/120 = 103/120.

Derived from ZIMSEC Statistics Paper 1, November 2020, Q1

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