AMC10 2002 A

AMC10 2002 A · Q24

AMC10 2002 A · Q24. It mainly tests Combinations, Probability (basic).

Tina randomly selects two distinct numbers from the set {1, 2, 3, 4, 5}, and Sergio randomly selects a number from the set {1, 2, ..., 10}. The probability that Sergio’s number is larger than the sum of the two numbers chosen by Tina is

Tina 从集合 {1, 2, 3, 4, 5} 中随机选取两个不同的数，Sergio 从 {1, 2, ..., 10} 中随机选取一个数。Sergio 的数大于 Tina 选取两个数之和的概率是

(A) 2/5 2/5

(B) 9/20 9/20

(D) 11/20 11/20

(E) 24/25 24/25

Answer

Correct choice: (A)

正确答案：（A）

Solution

(A) There are ten ways for Tina to select a pair of numbers. The sums 9, 8, 4, and 3 can be obtained in just one way, and the sums 7, 6, and 5 can each be obtained in two ways. The probability for each of Sergio’s choices is 1/10. Considering his selections in decreasing order, the total probability of Sergio’s choice being greater is \[ \left(\frac{1}{10}\right)\left(1+\frac{9}{10}+\frac{8}{10}+\frac{6}{10}+\frac{4}{10}+\frac{2}{10}+\frac{1}{10}+0+0+0\right)=\frac{2}{5}. \]

（A）Tina 选择一对数字共有十种方式。和为 9、8、4、3 的情况各只有一种方式得到，而和为 7、6、5 的情况各有两种方式得到。Sergio 每一种选择的概率都是 1/10。按从大到小考虑他的选择，Sergio 的选择更大的总概率为 \[ \left(\frac{1}{10}\right)\left(1+\frac{9}{10}+\frac{8}{10}+\frac{6}{10}+\frac{4}{10}+\frac{2}{10}+\frac{1}{10}+0+0+0\right)=\frac{2}{5}. \]

Topics