Independent solution

How to solve this Conditional Hypergeometric Counting question

Setup

Setup

Conditional on exactly six sales, all six-person subsets of the 12 customers are equally likely because the individual purchase probabilities are the same.

Nall=(126)N_{\mathrm{all}}=\binom{12}{6}

Model

Model

A favorable subset chooses two customers from each of the three interest groups. Multiply the three combination counts.

Nfav=(62)(42)(22)=90N_{\mathrm{fav}}=\binom62\binom42\binom22=90

Compute

Compute

There are 90 favorable subsets among 924 total six-person subsets, giving approximately 0.097403.

Pr⁡=90(126)=15154=0.097403\Pr=\frac{90}{\binom{12}{6}}=\frac{15}{154}=0.097403

Answer

Answer

The conditional composition probability is about 0.097.

0.097(D)\boxed{0.097\quad\text{(D)}}