Independent solution

How to solve this Counting Methods question

Answer in brief

A qualifying four-item subset must have category counts 2, 1, and 1. Counting the three possible doubled categories gives 105 favorable subsets out of 210, so the probability is 1/2 and choice E.

Setup

Setup

All four-item subsets of the ten available items are equally likely.

Nall=(104)=210N_{\mathrm{all}}=\binom{10}{4}=210

Model

Model

To include all three categories in a sample of four, exactly one category must contribute two items and each other category must contribute one.

Ngood=(52)(31)(21)+(51)(32)(21)+(51)(31)(22)N_{\mathrm{good}}=\binom{5}{2}\binom{3}{1}\binom{2}{1}+\binom{5}{1}\binom{3}{2}\binom{2}{1}+\binom{5}{1}\binom{3}{1}\binom{2}{2}

Compute

Compute

Evaluate the three disjoint category-count cases and divide by the total number of subsets.

Ngood=60+30+15=105N_{\mathrm{good}}=60+30+15=105
Pr⁡(all three categories)=105210=12\Pr(\text{all three categories})=\frac{105}{210}=\frac12

Answer

Answer

Exactly half of the equally likely four-item subsets contain every category.

12(E)\boxed{\frac12\quad\text{(E)}}