AMC12 2015 B

AMC12 2015 B · Q17

AMC12 2015 B · Q17. It mainly tests Linear equations, Probability (basic).

An unfair coin lands on heads with a probability of $\frac{1}{4}$. When tossed $n$ times, the probability of exactly two heads is the same as the probability of exactly three heads. What is the value of $n$?

一枚不公平的硬币正面朝上的概率为 $\frac{1}{4}$。抛掷 $n$ 次时，正好两次正面的概率等于正好三次正面的概率。$n$ 的值为多少？

(A) 5 5

(B) 8 8

(D) 11 11

(E) 13 13

Answer

Correct choice: (D)

正确答案：（D）

Solution

Answer (D): The probability of exactly two heads is $\left(\begin{array}{c} n \\ 2 \end{array}\right)\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^{n-2}$, and this must equal the probability of three heads, $\left(\begin{array}{c} n \\ 3 \end{array}\right)\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^{n-3}$. This results in the equation $$\frac{n(n-1)}{2}\cdot\frac{3}{4}=\frac{n(n-1)(n-2)}{6}\cdot\frac{1}{4}\quad \text{or}\quad \frac{3}{8}=\frac{n-2}{24}.$$ Therefore $n=11$.

答案（D）：恰好出现两个正面的概率是 $\left(\begin{array}{c} n \\ 2 \end{array}\right)\left(\frac{1}{4}\right)^2\left(\frac{3}{4}\right)^{n-2}$，且这必须等于出现三个正面的概率 $\left(\begin{array}{c} n \\ 3 \end{array}\right)\left(\frac{1}{4}\right)^3\left(\frac{3}{4}\right)^{n-3}$。因此得到方程 $$\frac{n(n-1)}{2}\cdot\frac{3}{4}=\frac{n(n-1)(n-2)}{6}\cdot\frac{1}{4}\quad \text{或}\quad \frac{3}{8}=\frac{n-2}{24}.$$ 因此 $n=11$。

Topics