阿米莉亞有一枚硬幣有$$ \frac{1}{3} $$概率落在正面，布萊恩有一枚硬幣有$$ \frac{2}{5} $$ 概率落在正面．阿米莉亞和布萊恩交替投擲硬幣，直到有人是正面；第一個是正面的人獲勝．所有拋硬幣的事件都是獨立的．阿米莉亞先拋，她獲勝的概率為$$ \frac{p}{q} $$，其中$$p$$和$$q$$是互質的正整數．$$q−p$$是多少？  Amelia has a coin that lands heads with probability$$ \frac{1}{3} $$ , and Blaine has a coin that lands on heads with probability$$ \frac{2}{5} $$ melia and Blaine alternately toss their coins untill 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 $$q---p$$?
选项为:A:$$1$$  B:$$2$$  C:$$3$$  D:$$4$$  E:$$5$$  请回答其中一个选项.