Question

Amelia has a coin that lands heads with probability $\frac{1}{3},$ and Blaine has a coin that lands on heads with probability $\frac{2}{5}.$ 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 $q-p?$

$1$

$2$

$3$

$4$

$5$

Solution:

Let $x$ be the probability that Amelia wins.

There is a $\frac{1}{3}$ chance Amelia wins off her first flip.

If she gets a tails, we want Blaine to lose, which happens with a $\frac{3}{5}$ chance.

The total probability of this case is $\dfrac{2}{3} \cdot \dfrac{3}{5} = \dfrac{2}{5}.$

The game then goes back to Amelia, who then again has a $x$ chance of winning.

Therefore, we get the following equation. $x = \dfrac{1}{3} + \dfrac{2}{5}x$ $\dfrac{3}{5}x = \dfrac{1}{3}$ $x = \dfrac{5}{9}.$

The difference between the denominator and numerator is $9 - 5 = 4.$

Thus, D is the correct answer.

2017 AMC 10A Problem 18

Solution:

Problem 18 in Other Years