AMC12 2010 A

AMC12 2010 A · Q15

AMC12 2010 A · Q15. It mainly tests Quadratic equations, Probability (basic).

A coin is altered so that the probability that it lands on heads is less than $\frac{1}{2}$ and when the coin is flipped four times, the probability of an equal number of heads and tails is $\frac{1}{6}$. What is the probability that the coin lands on heads?

一枚硬币被修改，使得它正面朝上的概率小于 $\frac{1}{2}$，当翻转四次时，正反面个数相等的概率为 $\frac{1}{6}$。这枚硬币正面朝上的概率是多少？

(A) $\frac{\sqrt{15}-3}{6}$ $\frac{\sqrt{15}-3}{6}$

(B) $\frac{6 - \sqrt{6+2}}{12}$ $\frac{6 - \sqrt{6+2}}{12}$

(D) $\frac{3 - \sqrt{3}}{6}$ $\frac{3 - \sqrt{3}}{6}$

(E) $\frac{\sqrt{3}-1}{2}$ $\frac{\sqrt{3}-1}{2}$

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let $x$ be the probability of flipping heads. It follows that the probability of flipping tails is $1-x$. The probability of flipping $2$ heads and $2$ tails is equal to the number of ways to flip it times the product of the probability of flipping each coin. \begin{align*}{4 \choose 2}x^2(1-x)^2 &= \frac{1}{6}\\ 6x^2(1-x)^2 &= \frac{1}{6}\\ x^2(1-x)^2 &= \frac{1}{36}\\ x(1-x) &= \pm\frac{1}{6}\end{align*} As for the desired probability $x$ both $x$ and $1-x$ are nonnegative, we only need to consider the positive root, hence \begin{align*}x(1-x) &= \frac{1}{6}\\ 6x^2-6x+1&=0\end{align*} Applying the quadratic formula we get that the roots of this equation are $\frac{3\pm\sqrt{3}}{6}$. As the probability of heads is less than $\frac{1}{2}$, we get that the answer is $\boxed{\textbf{(D)}\ \frac{3-\sqrt{3}}{6}}$.

设 $x$ 为出现正面的概率，则出现反面的概率为 $1-x$。出现 $2$ 次正面和 $2$ 次反面的概率等于出现这种结果的方式数乘以每次结果概率的乘积。 \begin{align*}{4 \choose 2}x^2(1-x)^2 &= \frac{1}{6}\\ 6x^2(1-x)^2 &= \frac{1}{6}\\ x^2(1-x)^2 &= \frac{1}{36}\\ x(1-x) &= \pm\frac{1}{6}\end{align*} 由于所求概率 $x$ 中，$x$ 与 $1-x$ 都非负，因此只需考虑正根，于是 \begin{align*}x(1-x) &= \frac{1}{6}\\ 6x^2-6x+1&=0\end{align*} 用求根公式可得方程的根为 $\frac{3\pm\sqrt{3}}{6}$。又因为正面概率小于 $\frac{1}{2}$，所以答案是 $\boxed{\textbf{(D)}\ \frac{3-\sqrt{3}}{6}}$。

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