1chancein 10,000 of being trapped in a much-maligned elevator in the mathematics building. if he goes to work 5 days a week, 52 weeks a year, for 10 years, and always rides the elevator up to his office when he first arrives, what is the probability that he will never be trapped? that he will be trapped once? twice? assume that the outcomes on all the days are mutually independent (a dubious assumption in practice)

Never being trapped = 77.10% Being trapped exactly once = 20.05% Being trapped exactly twice = 2.61% First, determine how many time the professor will ride the elevator, which will be 5 times 52 time 10 = 2600 times. The probability of his never being trapped will be (1-1/10000)^2600 = 0.771041562 = 77.1041562% which rounds to 77.1% Being trapped once is the probability of being trapped on just the 1st day, plus the probability of being trapped on just the second day, ..., plus the probability of being trapped on just the last day. So 0.0001 * 0.9999^2599 + 0.9999 * 0.0001 * 0.9999^2598 + 0.9999^2 * 0.0001 * 0.9999^2597 + ... + 0.9999^2599 * 0.0001 = (0.0001 * 0.9999^2599) * 2600 = 0.200490855 = 20.05% For being trapped exactly twice, the same type of calculation is done with the base value being 0.0001^2 * 0.9999^2598 The number of times the base value is multiplied will be the number of ways you can pick 2 days out of a set of 2600 days, which is 2600/(2!2598!) = 3378700 So we have (0.0001^2 * 0.9999^2598) * 3378700 = 0.026056392 = 2.61% The general formula for being trapped on exactly n days is 0.0001^n * 0.9999 ^ (2600-n) * 2600!/(n!(2600-n)!)

4.94.. let me know if i'm correct