1989 AIME 第 5 题

Question

将某枚不均匀的硬币抛掷五次，恰好出现一次正面的概率不为

0

，且与恰好出现两次正面的概率相同。设

5

次抛掷中恰好

3

次出现正面的概率化为最简分数后是

\frac{i}{j}

。求

i+j

。

When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to

0

and is the same as that of getting heads exactly twice. Let

\frac{i}{j},

in lowest terms, be the probability that the coin comes up heads in exactly

3

out of

5

flips. Find

i+j.

答案

设出现正面的概率为

p

，令题中的两个二项概率相等

Let

p

be the probability of heads and equate the two binomial probabilities

先约去非零公因式，再求

p

Cancel the nonzero common factors before solving for

p

设出现正面的概率为

p

。由条件可得

5p(1-p)^4=10p^2(1-p)^3\text{。}

题目所给的非零条件允许约分，得到

1-p=2p

，所以

p=\frac{1}{3}

。恰好出现三次正面的概率为

\binom53\left(\frac13\right)^3\left(\frac23\right)^2=\frac{40}{243}\text{。}

因此

i+j=40+243=283

。

Let

p

be the probability of heads. The condition gives

5p(1-p)^4=10p^2(1-p)^3.

The stated nonzero condition permits cancellation, yielding

1-p=2p,

so

p=\frac{1}{3}.

The probability of exactly three heads is

\binom53\left(\frac13\right)^3\left(\frac23\right)^2=\frac{40}{243}.

Therefore

i+j=40+243=283.

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