5.62 A certain airplane has two independent alternators to provide electrical power. The probability that a given alternator will fail on a 1-hour flight is .02. What is the probability that (a) both will fail? (b) Neither will fail? (c) One or the other will fail? Show all steps carefully.

P(An alternator will fail) = 0.02  P(An alternator will not fail) = 1 – 0.02 = 0.98
(a) P(Both will fail) = P(The first will fail and the second will fail) = 0.02 * 0.02 = 0.0004
(b) P(Neither will fail) = P(The first will not fail and the second will not fail)
= 0.98 * 0.98 = 0.9604
(c) P(One or the other will fail) = P(The first will fail) + P(The second will fail) – P(Both will fail)
= 0.02 + 0.02 – 0.0004 = 0.0396

5.70 The probability is 1 in 4,000,000 that a single auto trip in the United States will result in a fatality. Over a lifetime, an average U.S. driver takes 50,000 trips. (a) What is the probability of a fatal accident over a lifetime? Explain your reasoning carefully. Hint: Assume independent events. Why might the assumption of independence be violated? (b) Why might a driver be tempted not to use a seat belt “just on this trip”?

(a) If we assume independence, then we use a binomial distribution
n = 50000, p = 1/4000000 = 2.5 * 10^-7, q = 1 - p = 0.99999975
P(r fatalities in n trips) = nCr p^r q^(n - r)
P(One trip will be fatal) =[50000C1 (2.5 * 10^-7)^1 (0.99999975)^49999] = 0.012
The probability of meeting with a fatal accident = 0.012
The assumption of independence may be violated because in reality, the trips are not likely to be independent. For instance, most drivers get better with experience, so the probability of suffering a fatal accident should go down over time.
(b) The probability of a fatal accident in a single trip being as low as 1/4000000, a driver is likely to think that the odds of an accident are in his favor and therefore be tempted not to use a seat belt “just on this trip”.