Question

At a major International Airport, 0.35 of the flights arrive on time. A sample of 10 flights is studied. What is the probability that more than 3 of them arrived on time?

Write only a number as your answer. Round to 2 decimal places (for example 0.24). Do not write as a percentage.

Answer 1

p = 0.35

n = 10

It is a binomial distribution.

P(X = x) = nCx * px * (1 - p)n - x

P(X > 3) = 1 - P(X < 3)

              = 1 - (P(X = 0) + P(X = 1) + P(X = 2) = P(X = 3))

              = 1 - (10C0 * (0.35)^0 * (0.65)^10 + 10C1 * (0.35)^1 * (0.65)^9 + 10C2 * (0.35)^2 * (0.65)^8 + 10C3 * (0.35)^3 * (0.65)^7)

              = 1 - (0.01+0.07+0.18+0.25) = 0.49

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