2016 AMC 10B Problem 12

There are $\binom{5}{2} = 10$ equally likely pairs of different numbers.

A product is odd exactly when both factors are odd. The odd numbers available are $1, 3, 5$, giving $\binom{3}{2} = 3$ odd-odd pairs: $\{1,3\}, \{1,5\}, \{3,5\}$. Every other pair contains at least one even number and has an even product.

So $10 - 3 = 7$ pairs give an even product, and the probability is
$$\frac{7}{10} = 0.7.$$

The answer is $\boxed{\textbf{(D)}\ 0.7}$.

"Even product" means "at least one even factor," and "at least one" is a signal to count the complement, "no even factor," which is a single clean case. Parity of a product is decided by parity of the factors, so the actual values never matter, only how many are odd.

List the pairs containing an even number directly: pairs with $2$ or $4$. Pairs containing $2$: four of them; pairs containing $4$ but not $2$: three more. That is $7$ of $10$.

• Reasoning "two of the five numbers are even, so the probability is $2/5$," which ignores that a pair needs only one even factor.
• Counting ordered pairs ($20$) for the total but unordered pairs ($3$) for the odd case, giving a wrong fraction; keep both counts ordered or both unordered.