Question

Hey!

I'm beginning stats and probability and i came across this particular question. i don't understand it;

According to the National Centre for Health Statistics of the United States, in 2012, 15.6% of people under 65 years old had no health insurance coverage. If seven people under 65 years old were to be randomly selected, find the probability that at most two of them did not have health insurance.

Answer 1

$p=0.156 \\ q= 1-0.156 = 0.844 \\ n=7 \\ P(X=k) = C^n_k \times p^k \times q^{n-k} \\ P(X≤2) = P(X=0) + P(X=1) +P(X=2) \\ P(X=0) = \frac{7!}{0!(7-0)!} \times (0.156)^0 \times (0.844)^{7-0} \\ =1 \times 1 \times 0.3050 \\ = 0.3050 \\ P(X=1) = \frac{7!}{1!(7-1)!} \times (0.156)^1 \times (0.844)^{7-1} \\ = 7 \times 0.156 \times 0.3614 \\ = 0.3946 \\ P(X=2) = \frac{7!}{2!(7-2)!} \times (0.156)^2 \times (0.844)^{7-2} \\ = 21 \times 0.0243 \times 0.4282 \\ = 0.2185 \\ P(X≤2)=0.3050+0.3946+0.2185 = 0.9181$