Question

If a fair coin is tossed 9 times, what is the probability, rounded to the nearest thousandth, of getting at least 6 heads?

Answer 1

We need P(X ≥ 6) for X ~ Binomial(n=9, p=1/2):
P = [C(9,6)+C(9,7)+C(9,8)+C(9,9)] / 2^9 = (84+36+9+1)/512 = 130/512 = 65/256 ≈ 0.25390625.

Rounded to the nearest thousandth: 0.254.