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
Do you follow this step?

Answer 2

So Do I multiply 0.9 •0.9 -1?

Answer 3

3 airplanes so 3 is the denominator and the 2 planes that arrive on time then you have 2/3 chance

Answer 4

The solution is found as follows:
P(first two are on schedule and the third one is not on schedule) =
P(on schedule) * P(on schedule) * P(not on schedule) = you can calculate

Answer 5

Ok thanks

Answer 6

You're welcome :)