Question

An airline claims that 90% of the time, it's planes are on schedule. If three flights are selected at random , what is the probability that the first two are on schedule and the third one is not on schedule ?( please help me

Answer 1

P(on schedule) = 0.9
P(not on schedule) = 1.0 - 0.9 = 0.1
3 airplanes so 3 is the denominator and the 2 planes that arrive on time then you have 2/3 chance

P(first two are on schedule and the third one is not on schedule) = 0.9
P(on schedule) * P(on schedule) * P(not on schedule) = you can calculate
0.9 x 0.9 x 0.1 = 0.081