Question

A. According to an airline, flights on a certain route are independently of each other with an on-time rate of 80% of the time. Suppose 15 flights are randomly selected and the number of on time flights is recorded. How many total outcomes are available, if the order of the flights is considered?

B. According to an airline, flights on a certain route are independently of each other with an on-time rate of 80% of the time. Suppose 15 flights are randomly selected and the number of on time flights is recorded. What is the probability that more than 13 flights were on time?

C. According to an airline, flights on a certain route are independently of each other with an on-time rate of 80% of the time. Suppose 15 flights are randomly selected and the number of on time flights is recorded. What is the probability that less than 4 flights were not on time?

D. according to an airline, flights on a certain route are independently of each other with an on-time rate of 80% of the time. Suppose 15 flights are randomly selected and the number of on time flights is recorded. Given that more than 2 flights were late, what is the probability that at least 4 flights were late?

Answer 1

A) P = 0.8

n = 15

Total no of outcomes = np = 15 * 0.8 = 12

B) It is a binomial distribution.

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X > 13) = P(X = 14) + P(X = 15)

= 15C14 * (0.8)^14 * (0.2)^1 + 15C15 * (0.8)^15 * (0.2)^0 = 0.1671

C) P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= 15C0 * (0.2)^0 * (0.8)^15 + 15C1 * (0.2)^1 * (0.8)^14 + 15C2 * (0.2)^2 * (0.8)^13 + 15C3 * (0.2)^3 * (0.8)^12 = 0.6482

D) P(at least 4 flights were late | more than 2 flights were late) = P(at least 4 flights were late)/P(more than 2 flights were late)

= P(X > 4)/P(X > 2)

= (1 - P(X < 4))/(1 - P(X < 2))

= (1 - 0.6482)/(1 - (15C0 * (0.2)^0 * (0.8)^15 + 15C1 * (0.2)^1 * (0.8)^14 + 15C2 * (0.2)^2 * (0.8)^13 ))

= 0.3518/(1 - 0.3980)

= 0.3518/0.602

= 0.5844