Question

There are 10 flights daily from Bermuda via American Airlines into Newark International Airport.  Suppose the probability that any one of the flights arriving late is 0.10.  

What is the probability that more than 3 flights will be late today?

What is the probability that fewer than two flights will arrive late today?

What is the probability that 0 flights will arrive late today?

Answer 1

Here, n = 10, p = 0.1, (1 - p) = 0.9 and x =3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X >3).
P(X >3) = (10C4 * 0.1^4 * 0.9^6) + (10C5 * 0.1^5 * 0.9^5) + (10C6 * 0.1^6 * 0.9^4) + (10C7 * 0.1^7 * 0.9^3) + (10C8 * 0.1^8 * 0.9^2) + (10C9 * 0.1^9 * 0.9^1) + (10C10 * 0.1^10 * 0.9^0)
P(X >3) = 0.0112 + 0.0015 + 0.0001 + 0 + 0 + 0 + 0
P(X > 3) = 0.0128

b)

Here, n = 10, p = 0.1, (1 - p) = 0.9 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X < 2).
P(X < 2) = (10C0 * 0.1^0 * 0.9^10) + (10C1 * 0.1^1 * 0.9^9)
P(X < 2) = 0.3487 + 0.3874
P(X < 2) = 0.7361

c)

Here, n = 10, p = 0.1, (1 - p) = 0.9 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X = 0)
P(X = 0) = 10C0 * 0.1^0 * 0.9^10
P(X = 0) = 0.3487