1992 AMC 12 第 29 题

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

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

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)

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.

其他年份的第 29 题