2010 AMC 12A 第 15 题

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}

两次正面、两次反面的概率为

\binom{4}{2}p^2(1-p)^2

。

The chance of two heads and two tails is

\binom{4}{2}p^2(1-p)^2

这会化为

p(1-p)=\dfrac16

；解二次方程并取

p\lt\dfrac12

的根。

This reduces to

p(1-p)=\dfrac16;

solve the quadratic and take the root with

p\lt\dfrac12

设正面朝上的概率为

p

。抛掷四次恰好出现两次正面和两次反面的概率为

\begin{aligned} \binom{4}{2}p^2(1-p)^2 &= 6p^2(1-p)^2 \\ &= \frac16 \end{aligned}\text{。}

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.

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