AMC12 2005 B

AMC12 2005 B · Q11

AMC12 2005 B · Q11. It mainly tests Basic counting (rules of product/sum), 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?

一个信封里装有八张钞票：$2$ 张 1 美元、$2$ 张 5 美元、$2$ 张 10 美元和 $2$ 张 20 美元。随机不放回地抽取两张钞票。它们的总和为 $\$20$ 或更多的概率是多少？

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

(B) \frac{2}{7} \frac{2}{7}

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

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

Answer

Correct choice: (D)

正确答案：（D）

Solution

The only way to get a total of $20$ or more is if you pick a twenty and another bill, or if you pick both tens. There are a total of $\dbinom{8}{2}=\dfrac{8\times7}{2\times1}=28$ ways to choose $2$ bills out of $8$. There are $12$ ways to choose a twenty and some other non-twenty bill. There is $1$ way to choose both twenties, and also $1$ way to choose both tens. Adding these up, we find that there are a total of $14$ ways to attain a sum of $20$ or greater, so there is a total probability of $\dfrac{14}{28}=\boxed{\textbf{(D) }\frac{1}{2}}$.

要使总和达到 $20$ 或更多，唯一的方式是抽到一张 20 美元钞票和另一张钞票，或者抽到两张 10 美元钞票。总共有 $\dbinom{8}{2}=\dfrac{8\times7}{2\times1}=28$ 种从 $8$ 张钞票中选出 $2$ 张的方法。选出一张 20 美元钞票和另一张非 20 美元钞票有 $12$ 种方法。选出两张 20 美元钞票有 $1$ 种方法，选出两张 10 美元钞票也有 $1$ 种方法。将这些相加可得，总共有 $14$ 种方式使得总和达到 $20$ 或更大，因此概率为 $\dfrac{14}{28}=\boxed{\textbf{(D) }\frac{1}{2}}$。

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