Question

Assume that when adults with smartphones are randomly​ selected, 41​% use them in meetings or classes. If 3 adult smartphone users are randomly​ selected, find the probability that fewer than 3 of them use their smartphones in meetings or classes.

Answer 1

$p=0.41 \\ q=1-p = 1-0.41=0.59 \\ n = 3 \\ P(X=x) = \binom{n}{x}p^xq^{n-x} \\ P(X<3) = P(X=0) + P(X=1) +P(X=2) \\ P(X=0) = \frac{3!}{0!(3-0)!} \times 0.41^0 \times 0.59^{3-0} = 0.2053 \\ P(X=1) = \frac{3!}{1!(3-1)!} \times 0.41^1 \times 0.59^{3-1} = 0.4281 \\ P(X=2) = \frac{3!}{2!(3-2)!} \times 0.41^2 \times 0.59^{3-2} = 0.2975 \\ P(X<3) = 0.2053 + 0.4281 + 0.2975 = 0.9309$