Independent solution

How to solve this Binomial Distribution question

Setup

Setup

Let X be the defective count in a shipment and p the one-unit defect probability.

XBinomial(100,p)X\sim\operatorname{Binomial}(100,p)

Model

Model

Express the given comparison as a ratio of adjacent binomial masses.

Pr(X=3)Pr(X=2)=(1003)p3(1p)97(1002)p2(1p)98=2\frac{\Pr(X=3)}{\Pr(X=2)}=\frac{\binom{100}{3}p^3(1-p)^{97}}{\binom{100}{2}p^2(1-p)^{98}}=2

Compute

Compute

Cancel the common factors and solve the remaining linear equation.

983p1p=2\frac{98}{3}\frac{p}{1-p}=2
98p=6(1p),p=6104=352=0.057692307798p=6(1-p),\qquad p=\frac6{104}=\frac3{52}=0.0576923077

Answer

Answer

The defect probability rounds to 0.058.

0.058(C)\boxed{0.058\quad\text{(C)}}