Independent solution

How to solve this Binomial Distribution question

Setup

Setup

Let X be the number of awards among 20 independent cases, so X has a binomial distribution with success probability 0.02.

X∼Bin⁡(20,0.02)X\sim\operatorname{Bin}(20,0.02)
C=60⟹shortfall iff X≥3C=60\Longrightarrow \text{shortfall iff }X\ge3

Model

Model

For candidate payment C=60, the reserve supports two awards; a shortfall therefore occurs exactly when X is at least three.

Pr⁡(X≥3)=1−∑k=02(20k)(0.02)k(0.98)20−k\Pr(X\ge3)=1-\sum_{k=0}^{2}\binom{20}{k}(0.02)^k(0.98)^{20-k}

Compute

Compute

Use the complement of the binomial masses for zero, one, and two awards. The shortfall probability is 0.0070687, below the 0.01 limit.

Pr⁡(X≥3)=0.0070687<0.01\Pr(X\ge3)=0.0070687<0.01

Answer

Answer

Among the listed payment amounts, 60 is the largest that satisfies the shortfall constraint, so the answer is choice D.

C=60(D)\boxed{C=60\quad\text{(D)}}