Independent solution

How to solve this Binomial Distribution question

Setup

Setup

At most one event can occur in each year, independently, so the twenty-year count is binomial with event probability 0.05.

NBin(20,0.05)N\sim\operatorname{Bin}(20,0.05)

Model

Model

Fewer than three events includes the three disjoint counts zero, one, and two.

Pr(N<3)=k=02(20k)(0.05)k(0.95)20k\Pr(N<3)=\sum_{k=0}^{2}\binom{20}{k}(0.05)^k(0.95)^{20-k}

Compute

Compute

Adding those three binomial masses gives approximately 0.924516.

Pr(N<3)=0.924516\Pr(N<3)=0.924516

Answer

Answer

The probability of fewer than three events is approximately 0.92, selecting choice E.

0.92(E)\boxed{0.92\quad\text{(E)}}