Question

Jacob 用如下过程写出一个数列。首先他选择第一项为 $6$ 。为了生成下一项，他抛一枚公平硬币。若正面朝上，他将前一项加倍再减 $1$ 。若反面朝上，他取前一项的一半再减 $1$ 。Jacob 数列的第四项为整数的概率是多少？

Jacob uses the following procedure to write down a sequence of numbers. First he chooses the first term to be $6.$ To generate each succeeding term, he flips a fair coin. If it comes up heads, he doubles the previous term and subtracts $1.$ If it comes up tails, he takes half of the previous term and subtracts $1.$ What is the probability that the fourth term in Jacob's sequence is an integer?

$\dfrac{1}{6}$

$\dfrac{1}{3}$

$\dfrac{1}{2}$

$\dfrac{5}{8}$

$\dfrac{3}{4}$

答案：D

知识点：基本概率树状图奇偶性

难度评级：1880

解答：

从 $6$ 开始，第二项可能为 $11$ （正面）或 $2$ （反面）。

继续展开树形图，八个等可能的第四项为 $41, 9.5, 8, 1.25, 5, 0.5, -1, -1$ 。

其中 $41, 8, 5, -1, -1$ 是整数，所以概率为 $\dfrac{5}{8}$ 。

所以正确答案是 D。

Starting from $6,$ the second terms are $11$ (heads) and $2$ (tails).

Continuing the tree, the eight equally likely fourth terms are $41, 9.5, 8, 1.25, 5, 0.5, -1, -1.$

Of these, $41, 8, 5, -1, -1$ are integers, so the probability is $\dfrac{5}{8}.$

Thus, the correct answer is D.

2008 AMC 10A 第 22 题

答案：D

难度评级：1880

解答：

其他年份的第 22 题