2016 AMC 12A 第 19 题

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所有题目均经美国数学协会(MAA)官方合法授权使用。

19.

Jerry 从实数轴上的 00 出发。他投掷一枚公平硬币 88 次。若正面朝上,他向正方向移动 11 单位;若反面朝上,他向负方向移动 11 单位。他在这个过程中某一时刻到达 44 的概率为 ab\dfrac{a}{b},其中 aabb 是互质正整数。a+ba+b 是多少?(例如,若投掷序列为 HTHHHHHH,他成功。)

Jerry starts at 00 on the real number line. He tosses a fair coin 88 times. When he gets heads, he moves 11 unit in the positive direction; when he gets tails, he moves 11 unit in the negative direction. The probability that he reaches 44 at some time during this process is ab,\dfrac{a}{b}, where aa and bb are relatively prime positive integers. What is a+b?a+b? (For example, he succeeds if his sequence of tosses is HTHHHHHH.)

6969

151151

257257

293293

313313

答案:B
知识点:随机游走分类讨论
难度评级:1990
小提示:

统计 282^8 个投掷序列中,累计位置到达 44 的序列数。

Count the 282^8 toss sequences in which the running total reaches 44

大提示:

反面至多 22 次时一定到达 44;反面有 3344 次时只有部分顺序可行。

With at most 22 tails he always reaches 4;4; with 33 or 44 tails only some orders work

解答:

统计 88 次投掷中累计总和达到过 44 的序列数。如果反面至多出现 22 次,他一定能达到 44,贡献 (80)+(81)+(82)=1+8+28=37 \begin{gathered} \binom80+\binom81+\binom82\\ =1+8+28\\ =37 \end{gathered} 个序列。

恰有 33 次反面时,他只能在第 44 次或第 66 次投掷首次达到 44。要在第 44 次达到,前四次必须全为正面,之后剩下的那个正面有 44 个可能位置。否则第 55 和第 66 次是正面,前四次中有一次反面,最后两次是反面,同样有 44 种可能。因此这一情形贡献 88 个序列。恰有 44 次反面时,只有 HHHHTTTT 可行,贡献 11 个序列。正面少于 44 次时不可能达到 44

所以有利序列共有 37+8+1=4637+8+1=46 个,全部序列有 28=2562^8=256 个,概率为 46256=23128\dfrac{46}{256}=\dfrac{23}{128}。因此 a+b=23+128=151a+b=23+128=151

因此,正确答案是 B

Count the sequences of 88 tosses whose running total reaches 4.4. With at most 22 tails he certainly reaches 4,4, contributing (80)+(81)+(82)=1+8+28=37 \begin{gathered} \binom80+\binom81+\binom82\\ =1+8+28\\ =37 \end{gathered} sequences.

With exactly 33 tails, he can first reach 44 on toss 44 or toss 6.6. Reaching it on toss 44 requires four initial heads, after which the remaining head has 44 possible positions. Otherwise, tosses 55 and 66 are heads, one of the first four tosses is a tail, and the last two are tails, again giving 44 possibilities. Thus this case contributes 8.8. With exactly 44 tails, only HHHHTTTT works, giving 1.1. He cannot reach 44 with fewer than 44 heads.

So there are 37+8+1=4637+8+1=46 favorable sequences out of 28=256,2^8=256, a probability of 46256=23128.\dfrac{46}{256}=\dfrac{23}{128}. Then a+b=23+128=151.a+b=23+128=151.

Thus, the correct answer is B.

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