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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
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Jan 28, 2017
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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
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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
MaZZZu
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Feb 2, 2017
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