Question

28.1% of adults have more than 2 cars. Suppose that a group of 30 adults were chosen at random. Find the probability that at least 10 of those adults have more than 2 cars

Answer 1

Here, n = 30, p = 0.281, (1 - p) = 0.719 and x = 10
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 9).
P(X <= 9) = (30C0 * 0.281^0 * 0.719^30) + (30C1 * 0.281^1 * 0.719^29) + (30C2 * 0.281^2 * 0.719^28) + (30C3 * 0.281^3 * 0.719^27) + (30C4 * 0.281^4 * 0.719^26) + (30C5 * 0.281^5 * 0.719^25) + (30C6 * 0.281^6 * 0.719^24) + (30C7 * 0.281^7 * 0.719^23) + (30C8 * 0.281^8 * 0.719^22) + (30C9 * 0.281^9 * 0.719^21)
P(X <= 9) = 0.0001 + 0.0006 + 0.0033 + 0.0122 + 0.0322 + 0.0654 + 0.1065 + 0.1427 + 0.1603 + 0.1532
P(X <= 9) = 0.6765

P(X >= 10) = 1 - P(x< =9)
= 1 - 0.6765
= 0.3235