AMC10 2005 B

AMC10 2005 B · Q15

AMC10 2005 B · Q15. It mainly tests Combinations, Probability (basic).

An envelope contains eight bills: 2 ones, 2 fives, 2 tens, and 2 twenties. Two bills are drawn at random without replacement. What is the probability that their sum is $20$ or more?

一个信封里有 8 张钞票：2 张 1 元，2 张 5 元，2 张 10 元，2 张 20 元。随机不放回抽取 2 张钞票。它们的和为 20 元或更多的概率是多少？

(A) $\frac{1}{4}$ $\frac{1}{4}$

(B) $\frac{2}{5}$ $\frac{2}{5}$

(D) $\frac{1}{2}$ $\frac{1}{2}$

(E) $\frac{2}{3}$ $\frac{2}{3}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) There are \[ \binom{8}{2}=\frac{8!}{6!\cdot 2!}=28 \] ways to choose the bills. A sum of at least \$20 is obtained by choosing both \$20 bills, one of the \$20 bills and one of the six smaller bills, or both \$10 bills. Hence the probability is \[ \frac{1+2\cdot 6+1}{28}=\frac{14}{28}=\frac{1}{2}. \]

(D) 共有 \[ \binom{8}{2}=\frac{8!}{6!\cdot 2!}=28 \] 种选取钞票的方法。至少得到 \$20 的总金额可以通过选取两张 \$20 钞票、选取一张 \$20 钞票并再选取六张较小面额钞票中的一张，或者选取两张 \$10 钞票来实现。因此概率为 \[ \frac{1+2\cdot 6+1}{28}=\frac{14}{28}=\frac{1}{2}. \]

Topics