2016 AMC 10B Problem 12

1. Count all the pairs. From $1, 2, 3, 4, 5$ there are $10$ ways to pick two different numbers:
$$\{1,2\}, \{1,3\}, \{1,4\}, \{1,5\}, \{2,3\}, \{2,4\}, \{2,5\}, \{3,4\}, \{3,5\}, \{4,5\}.$$
2. When is a product even? If even one of the two numbers is even, the product is even. The product is odd only when both numbers are odd.
3. Counting the odd products is easier, so do that. The odd numbers are $1, 3, 5$, and the pairs using only them are $\{1,3\}, \{1,5\}, \{3,5\}$. That is $3$ pairs.
4. The other $10 - 3 = 7$ pairs give an even product.
5. The probability is $\frac{7}{10} = 0.7$.

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

"Even product" means "at least one even number." When a question says "at least one," it is often easier to count the opposite ("none") and subtract. Here the opposite is "both odd," which has only three pairs.

Count the even products directly. Pairs that include $2$: $\{1,2\}, \{2,3\}, \{2,4\}, \{2,5\}$, which is $4$ pairs. Pairs that include $4$ but not $2$: $\{1,4\}, \{3,4\}, \{4,5\}$, which is $3$ more. That is $7$ out of $10$.

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 $\boxed{\textbf{(D)}\ 0.7}$.