This item requires multiplying conditional probabilities and then adding mutually exclusive outcomes. “Taking the bus and being on time” means the commuter takes the bus and is not late. Since the probability of being late by bus is .15, the probability of being on time by bus is 1 − .15 = .85. Therefore, P(bus and on time) = .4 × .85 = .34. For driving, the probability of being late is .05, so the probability of being on time is 1 − .05 = .95. Therefore, P(car and on time) = .4 × .95 = .38. Because a commuter cannot both take the bus and drive a car on the same trip, the events are mutually exclusive. Add the two joint probabilities: .34 + .38 = .72. The cycling information is not used because the question asks only about bus-on-time or car-on-time outcomes. References/topics from the Study Guide: probability rules, complements, conditional probability, mutually exclusive events.

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