AMC10 2004 A

AMC10 2004 A · Q10

AMC10 2004 A · Q10. It mainly tests Basic counting (rules of product/sum), Probability (basic).

Coin A is flipped three times and coin B is flipped four times. What is the probability that the number of heads obtained from flipping the two fair coins is the same?

硬币 A 抛三次，硬币 B 抛四次。抛两个公平硬币得到的正面次数相同的概率是多少？

(A) 19/128 $\frac{19}{128}$

(B) 23/128 $\frac{23}{128}$

(D) 35/128 $\frac{35}{128}$

(E) 1/2 $\frac{1}{2}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) The result will occur when both $A$ and $B$ have either 0, 1, 2, or 3 heads, and these probabilities are shown in the table. \[ \begin{array}{c|cccc} \text{Heads} & 0 & 1 & 2 & 3\\ \hline A & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8}\\ \hline B & \frac{1}{16} & \frac{4}{16} & \frac{6}{16} & \frac{4}{16} \end{array} \] The probability of both coins having the same number of heads is \[ \frac{1}{8}\cdot\frac{1}{16}+\frac{3}{8}\cdot\frac{4}{16}+\frac{3}{8}\cdot\frac{6}{16}+\frac{1}{8}\cdot\frac{4}{16}=\frac{35}{128}. \]

（D）当 $A$ 和 $B$ 都出现 0、1、2 或 3 个正面时，该结果会发生，这些概率如下表所示。 \[ \begin{array}{c|cccc} \text{正面数} & 0 & 1 & 2 & 3\\ \hline A & \frac{1}{8} & \frac{3}{8} & \frac{3}{8} & \frac{1}{8}\\ \hline B & \frac{1}{16} & \frac{4}{16} & \frac{6}{16} & \frac{4}{16} \end{array} \] 两枚硬币出现相同正面数的概率为 \[ \frac{1}{8}\cdot\frac{1}{16}+\frac{3}{8}\cdot\frac{4}{16}+\frac{3}{8}\cdot\frac{6}{16}+\frac{1}{8}\cdot\frac{4}{16}=\frac{35}{128}. \]

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