两素数之和为偶数的概率

1..20 的素数为 $2,3,5,7,11,13,17,19$ 共 8 个，其中只有 2 是偶数。

两数之和为偶数当且仅当两数同奇偶。由于只能有一个偶素数 2，而题目要求两数不同，因此不可能选到两个偶数。

所以必须选到两个奇素数（即不选 2）。

概率为

$$\frac{\binom{7}{2}}{\binom{8}{2}}=\frac{21}{28}=\frac{3}{4}.$$

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Original Explanation

The prime integers at most $20$ are $2, 3, 5, 7, 11, 13, 17,$ and $19$, which yields $8$ total integers. There are $8 \cdot 7 = 56$ ways to pick $2$ integers ordered for the sum. Our sum is even exactly when we don't select $2$. Therefore, there are $7 \cdot 6 = 42$ ways to pick two prime integers that aren't $2$. Therefore, our probability is

$$\frac{42}{56} = \frac{3}{4}$$