Question

一枚“不均匀”硬币出现正面的概率为 $\frac23$ 。若将这枚硬币抛掷 $50$ 次，正面出现总次数为偶数的概率是多少？

An “unfair” coin has a $\frac23$ probability of turning up heads. If this coin is tossed $50$ times, what is the probability that the total number of heads is even?

$2^5\left(\frac23\right)^{50}$

$\frac12\left(1-\frac1{3^{50}}\right)$

$\frac12$

$\frac12\left(1+\frac1{3^{50}}\right)$

$\frac23$

答案：D

知识点：binomial parity probability recurrence complementary states

难度评级：2310

小提示：

追踪正面次数为偶数与为奇数的概率之差

Track the difference between the probabilities of an even and an odd number of heads

大提示：

每抛掷一次，这个差会乘以 $P(T)-P(H)$

One toss multiplies that difference by $P(T)-P(H)$

解答：

设 $E_n$ 和 $O_n$ 分别为抛掷 $n$ 次后正面次数为偶数和奇数的概率。则 $E_n+O_n=1$ ，并且 $\begin{aligned} E_{n+1}-O_{n+1} &=\left(\frac13-\frac23\right)\\ &\qquad\cdot(E_n-O_n) \end{aligned}\text{。}$ 因为 $E_0-O_0=1$ ，所以 $E_{50}-O_{50}=(-\frac{1}{3})^{50}=3^{-50}$ 。联立和与差的方程，得到 $E_{50}=\frac12\left(1+\frac1{3^{50}}\right)\text{。}$

因此正确答案是 D。

Let $E_n$ and $O_n$ be the probabilities of an even and odd number of heads after $n$ tosses. Then $E_n+O_n=1,$ while $\begin{aligned} E_{n+1}-O_{n+1} &=\left(\frac13-\frac23\right)\\ &\qquad\cdot(E_n-O_n). \end{aligned}$ Since $E_0-O_0=1,$ we have $E_{50}-O_{50}=(-\frac{1}{3})^{50}=3^{-50}.$ Solving the sum and difference equations gives $E_{50}=\frac12\left(1+\frac1{3^{50}}\right).$

Thus the correct answer is D.

1992 AMC 12 第 29 题

答案：D

难度评级：2310

小提示：

大提示：

解答：

其他年份的第 29 题