Answer to Question #337528 in Statistics and Probability for Andrew

Question

A coin is tossed 10 times. Find the probability that at least three are tails.

Answer 1

Let $X=$ the number of tails: $X\sim Bin (n, p).$

Given $p=0.5, q=1-p=0.5,n=10.$

$$P(X\ge 3)=1-P(X=0)-P(X=1)$$

$$-P(X=2)=1-\dbinom{10}{0}(0.5)^0(0.5)^{10-0}$$

$$-\dbinom{10}{1}(0.5)^1(0.5)^{10-1}-\dbinom{10}{2}(0.5)^2(0.5)^{10-2}$$

$$=0.9453125$$