Try this beautiful Problem on Probability based on Coin toss from AMC 10 A, 2017. You may use sequential hints to solve the problem.

## Coin Toss – AMC-10A, 2017- Problem 18

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 $q-p ?$

,

- $1$
- $2$
- $3$
- $4$
- $5$

**Key Concepts**

combinatorics

Coin toss

Probability

## Suggested Book | Source | Answer

#### Suggested Reading

Pre College Mathematics

#### Source of the problem

AMC-10A, 2017 Problem-18

#### Check the answer here, but try the problem first

$4$

## Try with Hints

#### First Hint

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.

Now can you finish the problem?

#### Second Hint

Let $P$ be the probability Amelia wins. Note that $P=$ chance she wins on her first turn $+$ chance she gets to her second turn $\cdot \frac{1}{3}+$ chance she gets to her third turn $\cdot \frac{1}{3} \ldots$ This can be represented by an infinite geometric series,

Therefore the value of \(P\) will be $P=\frac{\frac{1}{3}}{1-\frac{2}{3} \cdot \frac{3}{5}}=\frac{\frac{1}{3}}{1-\frac{2}{5}}=\frac{\frac{1}{3}}{\frac{3}{5}}=\frac{1}{3} \cdot \frac{5}{3}=\frac{5}{9}$ which is of the form \(\frac{p}{q}\)

Now Can you finish the Problem?

#### Third Hint

Therefore \(q-p=9-5=4\)

## Other useful links

- https://www.cheenta.com/surface-area-of-cube-amc-10a-2007-problem-21/
- https://www.youtube.com/watch?v=h4MmDUky7KM

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