Question

一枚硬币被改造后，正面朝上的概率小于 $\dfrac12$ ，抛这枚硬币四次，正面和反面次数相等的概率为 $\dfrac{1}{6}$ 。这枚硬币正面朝上的概率是多少？

A coin is altered so that the probability that it lands on heads is less than $\dfrac12,$ and when the coin is flipped four times, the probability of an equal number of heads and tails is $\dfrac{1}{6}.$ What is the probability that the coin lands on heads?

$\dfrac{\sqrt{15}-3}{6}$

$\dfrac{6-\sqrt{6\sqrt6+2}}{12}$

$\dfrac{\sqrt2-1}{2}$

$\dfrac{3-\sqrt3}{6}$

$\dfrac{\sqrt3-1}{2}$

答案：D

知识点：二项概率二次方程

难度评级：1650

解答：

设正面朝上的概率为 $p$ 。 $\begin{aligned} \binom{4}{2}p^2(1-p)^2 &= 6p^2(1-p)^2 \\ &= \frac16. \end{aligned}$

因此 $p^2(1-p)^2=\dfrac{1}{36}$ ，所以 $p(1-p)=\dfrac16$ 。

这给出 $6p^2-6p+1=0$ ，故 $p=\dfrac{3\pm\sqrt3}{6}$ 。由于 $p\lt\dfrac12$ ，取 $p=\dfrac{3-\sqrt3}{6}$ 。

因此 D 是正确答案。

Let $p$ be the probability of heads. The chance of two heads and two tails in four flips is $\begin{aligned} \binom{4}{2}p^2(1-p)^2 &= 6p^2(1-p)^2 \\ &= \frac16. \end{aligned}$

Thus $p^2(1-p)^2=\dfrac{1}{36},$ so $p(1-p)=\dfrac16.$

This gives $6p^2-6p+1=0,$ so $p=\dfrac{3\pm\sqrt3}{6}.$ Since $p\lt\dfrac12,$ we take $p=\dfrac{3-\sqrt3}{6}.$

Thus, D is the correct answer.

2010 AMC 12A 第 15 题

答案：D

难度评级：1650

解答：

其他年份的第 15 题