An article reported that what airline passengers like to do most on long flights is rest or sleep; in a survey of 3697 passengers, almost 90% did so. Suppose that for a particular route the actual percentage is exactly 90%, and consider randomly selecting six passengers. Then x, the number among the selected six who rested or slept, is a binomial random variable with n = 6 and p = 0.9. (Round your answers to four decimal places.)

(a) Calculate p(4). p(4) = Interpret this probability. This is the probability that at least 4 out of 10 selected passengers rested or slept. This is the probability that exactly 4 out of 6 selected passengers rested or slept. This is the probability that exactly 4 out of 10 selected passengers rested or slept. This is the probability that at least 4 out of 6 selected passengers rested or slept.
(b) Calculate p(6), the probability that all six selected passengers rested or slept. p(6) =
(c) Determine P(x ≥ 4). P(x ≥ 4) =