AMC10 2011 A

AMC10 2011 A · Q21

AMC10 2011 A · Q21. It mainly tests Basic counting (rules of product/sum), Casework.

Two counterfeit coins of equal weight are mixed with 8 identical genuine coins. The weight of each of the counterfeit coins is different from the weight of each of the genuine coins. A pair of coins is selected at random without replacement from the 10 coins. A second pair is selected at random without replacement from the remaining 8 coins. The combined weight of the first pair is equal to the combined weight of the second pair. What is the probability that all 4 selected coins are genuine?

有两个重量相等的假币与8个相同的真币混合在一起。每个假币的重量与每个真币的重量不同。从这10枚硬币中随机不放回地选出一对硬币。从剩余的8枚硬币中随机不放回地选出第二对硬币。第一对硬币的总重量等于第二对硬币的总重量。4枚选出的硬币全为真币的概率是多少？

(A) $\frac{7}{11}$ $\frac{7}{11}$

(B) $\frac{9}{13}$ $\frac{9}{13}$

(D) $\frac{15}{19}$ $\frac{15}{19}$

(E) $\frac{15}{16}$ $\frac{15}{16}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The weights of the two pairs of coins are equal if each pair contains the same number of counterfeit coins. Therefore either the first pair and the second pair both contain only genuine coins, or the first pair and the second pair both contain one counterfeit coin. The number of ways to choose the coins in the first case is $\binom{8}{2}\cdot\binom{6}{2}=420$. The number of ways to choose the coins in the second case is $8\cdot2\cdot7\cdot1=112$. Therefore the requested probability is $\frac{420}{112+420}=\frac{15}{19}$.

答案（D）：当每一对硬币中假币的数量相同的时候，两对硬币的重量相等。因此，要么第一对和第二对都只包含真币，要么第一对和第二对都各包含一枚假币。第一种情况选硬币的方法数为 $\binom{8}{2}\cdot\binom{6}{2}=420$。第二种情况选硬币的方法数为 $8\cdot2\cdot7\cdot1=112$。因此所求概率为 $\frac{420}{112+420}=\frac{15}{19}$。

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