AMC10 2018 B

AMC10 2018 B · Q6

AMC10 2018 B · Q6. It mainly tests Basic counting (rules of product/sum), Probability (basic).

A box contains 5 chips, numbered 1, 2, 3, 4, and 5. Chips are drawn randomly one at a time without replacement until the sum of the values drawn exceeds 4. What is the probability that 3 draws are required?

一个盒子里有5个标有数字1、2、3、4和5的筹码。随机依次不放回地抽取筹码，直到抽取的数字之和超过4。需要3次抽取的概率是多少？

(A) $\frac{1}{15}$ $\frac{1}{15}$

(B) $\frac{1}{10}$ $\frac{1}{10}$

(D) $\frac{1}{5}$ $\frac{1}{5}$

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

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): Three draws will be required if and only if the first two chips drawn have a sum of 4 or less. The draws $(1,2)$, $(2,1)$, $(1,3)$, and $(3,1)$ are the only draws meeting this condition. There are $5\cdot 4=20$ possible two-chip draws, so the requested probability is $\frac{4}{20}=\frac{1}{5}$. (Note that all 20 possible two-chip draws are considered in determining the denominator, even though some draws will end after the first chip is drawn.)

答案（D）：当且仅当前两次抽到的筹码之和不超过 $4$ 时，才需要抽第三次。满足该条件的抽取结果只有 $(1,2)$、$(2,1)$、$(1,3)$ 和 $(3,1)$。共有 $5\cdot 4=20$ 种可能的两次抽取结果，因此所求概率为 $\frac{4}{20}=\frac{1}{5}$。（注意：在确定分母时，会把全部 20 种两次抽取结果都计入，尽管有些情况下在抽到第一枚筹码后过程就会结束。）

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