Independent solution

How to solve this Hypergeometric Distribution question

Setup

Setup

There are 120 equally likely three-house damage sets. Express the all-uninsured event using k.

(103)=120\binom{10}{3}=120
(k3)(103)=1120\frac{\binom{k}{3}}{\binom{10}{3}}=\frac1{120}

Model

Model

The probability equation determines the number of uninsured and insured houses.

(k3)=1⟹k=3\binom{k}{3}=1\quad\Longrightarrow\quad k=3
10−k=710-k=7

Compute

Compute

At most one insured house means either zero insured and three uninsured, or one insured and two uninsured.

Pr⁡(I≤1)=(70)(33)+(71)(32)(103)\Pr(I\le1)=\frac{\binom70\binom33+\binom71\binom32}{\binom{10}{3}}
Pr⁡(I≤1)=1+21120=1160\Pr(I\le1)=\frac{1+21}{120}=\frac{11}{60}

Answer

Answer

The probability is 11/60.

1160(C)\boxed{\frac{11}{60}\quad\text{(C)}}