AMC10 2017 A

AMC10 2017 A · Q18

AMC10 2017 A · Q18. It mainly tests Probability (basic), Conditional probability (basic).

Amelia has a coin that lands on heads with probability \(\frac{1}{3}\), and Blaine has a coin that lands on heads with probability \(\frac{2}{5}\). Amelia and Blaine alternately toss their coins until someone gets a head; the first one to get a head wins. All coin tosses are independent. Amelia goes first. The probability that Amelia wins is \(\frac{p}{q}\), where p and q are relatively prime positive integers. What is \(q - p\)?

Amelia有一枚正面朝上的概率为\(\frac{1}{3}\)的硬币，Blaine有一枚正面朝上的概率为\(\frac{2}{5}\)的硬币。Amelia和Blaine轮流抛硬币，直到有人得到正面；第一个得到正面的人获胜。所有抛硬币是独立的。Amelia先抛。Amelia获胜的概率为\(\frac{p}{q}\)。其中p和q互质正整数。求\(q - p\)？

(A) 1 1

(B) 2 2

(D) 4 4

(E) 5 5

Answer

Correct choice: (D)

正确答案：（D）

Solution

Let \(x\) be the probability that Amelia wins. Then\n\[x = \frac{1}{3} + \left(1 - \frac{1}{3}\right)\left(1 - \frac{2}{5}\right)x,\]\nbecause either Amelia wins on the first toss, or, if she and Blaine both get tails, then the chance of her winning from that point onward is also \(x\). Solving this equation gives \(x = \frac{5}{9}\). The requested difference is \(9 - 5 = 4\).

令\(x\)为Amelia获胜的概率。那么\n\[x = \frac{1}{3} + \left(1 - \frac{1}{3}\right)\left(1 - \frac{2}{5}\right)x,\]\n因为要么Amelia第一次抛就赢，要么如果她和Blaine都得到反面，那么从那时起她获胜的概率仍是\(x\)。解此方程得\(x = \frac{5}{9}\)。要求的差值为\(9 - 5 = 4\)。

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