Suppose that last month, WestJet flights arrived on time 77% of the time and you were on 3 of those flights. (Round to 4 decimal places if needed)

Question

a) Given that each flight is independent of one another, what is the probability that all of your flights arrived on time?

b) What is the probability that none of your flights arrived on time?

c) What is the probability that at least one of your flights arrived on time?

Answer 1

These are all special cases of the "Binomial Distribution".

a) Given that each flight is independent of one another, what is the probability that all of your flights arrived on time?

P(all on time) = P(1st_on_time & 2nd_on_time & 3rd_on_time & 4th_on_time)

Because each flight is independent, we can express this as a product.

P(all on time) = P(1st_on_time) * P(2nd_on_time) * P(3rd_on_time) * P(4th_on_time)

Each event is identically distributed. (Look up "independent identically distributed random variables" and "Bernoulli distribution")

P(all on time) = 0.77 * 0.77 * 0.77 * 0.77

P(all on time) = 0.77^4

P(all on time) = 0.3515304

P(X=4) = 0.3515

b) What is the probability that none of your flights arrived on time?

Look up "complementary events". Each flight is either on time or not. If there is a 77% chance on on-time, then there is a 23% chance of not (for each flight).

P(none on time) = (1-0.77)^4 = 0.23^4 = 0.00279841

P(X=0) = 0.0028

c) What is the probability that at least one of your flights arrived on time?

This entire event (X>0) is the complement of the previous event (X=0). This is because it is impossible for the number of on-time flights to equal anything but an integer between 0 and 4 (inclusive).

Event[X>0] is equivalent to Event[NOT X=0]

P(X>0) = 1-0.0028 = 0.9972