AMC12 2016 A

AMC12 2016 A · Q19

AMC12 2016 A · Q19. It mainly tests Probability (basic), Casework.

Jerry starts at 0 on the real number line. He tosses a fair coin 8 times. When he gets heads, he moves 1 unit in the positive direction; when he gets tails, he moves 1 unit in the negative direction. The probability that he reaches 4 at some time during this process is $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. What is $a+b$? (For example, he succeeds if his sequence of tosses is HTHHHHHH.)

杰瑞从实数轴上的 0 开始。他掷一枚公平硬币 8 次。每次掷出正面就向正方向移动 1 个单位；每次掷出反面就向负方向移动 1 个单位。在这一过程中，他在某个时刻到达 4 的概率是 $\frac{a}{b}$，其中 $a$ 和 $b$ 是互质的正整数。求 $a+b$。（例如，如果他的掷币序列是 HTHHHHHH，则他成功。）

(A) 69 69

(B) 151 151

(D) 293 293

(E) 313 313

Answer

Correct choice: (B)

正确答案：（B）

Solution

Answer (B): Jerry arrives at 4 for the first time after an even number of tosses. Because Jerry tosses 8 coins, he arrives at 4 for the first time after either 4, 6, or 8 tosses. If Jerry arrives at 4 for the first time after 4 tosses, then he must have tossed HHHH. The probability of this occurring is $\frac{1}{16}$. If Jerry arrives at 4 for the first time after 6 tosses, he must have tossed 5 heads and 1 tail among the 6 tosses, and the 1 tail must have come among the first 4 tosses. Thus, there are 4 possible sequences of valid tosses, each with probability $\frac{1}{64}$, for a total of $\frac{4}{64}=\frac{1}{16}$. If Jerry arrives at 4 for the first time after 8 tosses, then he must have tossed 6 heads and 2 tails among the 8 tosses. Both tails must occur among the first 6 tosses; otherwise Jerry would have already reached 4 before the 8th toss. Further, at least 1 tail must occur in the first 4 tosses; otherwise Jerry would have already reached 4 after the 4th toss. Therefore there are $\binom{6}{2}-1=14$ sequences for which Jerry first arrives at 4 after 8 tosses, each with probability $\frac{1}{256}$, for a total of $\frac{14}{256}=\frac{7}{128}$. Thus the probability that Jerry reaches 4 at some time during the process is $\frac{1}{16}+\frac{1}{16}+\frac{7}{128}=\frac{23}{128}$. The requested sum is $23+128=151$.

答案（B）：Jerry 第一次到达 4 一定发生在抛掷次数为偶数之后。因为 Jerry 一共抛 8 次硬币，所以他第一次到达 4 只能发生在抛了 4 次、6 次或 8 次之后。若 Jerry 第一次在抛 4 次后到达 4，则他必须抛出 HHHH，其概率为 $\frac{1}{16}$。若 Jerry 第一次在抛 6 次后到达 4，则 6 次中必须有 5 次正面、1 次反面，并且这 1 次反面必须出现在前 4 次之中。因此共有 4 种符合条件的抛掷序列，每种概率为 $\frac{1}{64}$，总概率为 $\frac{4}{64}=\frac{1}{16}$。若 Jerry 第一次在抛 8 次后到达 4，则 8 次中必须有 6 次正面、2 次反面。两次反面都必须出现在前 6 次之中；否则 Jerry 会在第 8 次之前就已经到达 4。此外，前 4 次中至少要出现 1 次反面；否则 Jerry 会在第 4 次后就已经到达 4。因此，使得 Jerry 第一次在抛 8 次后到达 4 的序列数为 $\binom{6}{2}-1=14$，每种概率为 $\frac{1}{256}$，总概率为 $\frac{14}{256}=\frac{7}{128}$。所以 Jerry 在过程中某个时刻到达 4 的概率为 $\frac{1}{16}+\frac{1}{16}+\frac{7}{128}=\frac{23}{128}$。所求的和为 $23+128=151$。

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