Difference between revisions of "2017 AMC 10A Problems/Problem 18"

m (Solution 2)
m
 
(One intermediate revision by one other user not shown)
Line 16: Line 16:
  
 
~minor LaTeX edit by virjoy2001
 
~minor LaTeX edit by virjoy2001
 +
 +
== Video Solution ==
 +
https://youtu.be/IRyWOZQMTV8?t=4552
 +
 +
~ pi_is_3.14
  
 
==Video Solution==
 
==Video Solution==
Line 24: Line 29:
 
{{MAA Notice}}
 
{{MAA Notice}}
  
[[Category:Intermediate Probability Problems]]
+
[[Category:Introductory Probability Problems]]

Latest revision as of 12:37, 10 April 2021

Problem

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 $q-p$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

Solution 1

Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her turn again}\cdot P$, as if she gets to her turn again, she is back where she started with probability of winning $P$. The chance she wins on her first turn is $\frac{1}{3}$. The chance she makes it to her turn again is a combination of her failing to win the first turn - $\frac{2}{3}$ and Blaine failing to win - $\frac{3}{5}$. Multiplying gives us $\frac{2}{5}$. Thus, \[P = \frac{1}{3} + \frac{2}{5}P\] Therefore, $P = \frac{5}{9}$, so the answer is $9-5=\boxed{\textbf{(D)}\ 4}$.

Solution 2

Let $P$ be the probability Amelia wins. Note that $P = \text{chance she wins on her first turn} + \text{chance she gets to her second turn}\cdot \frac{1}{3} + \text{chance she gets to her third turn}\cdot \frac{1}{3} ...$This can be represented by an infinite geometric series: \[P=\frac{\frac{1}{3}}{1-\frac{2}{3}\cdot \frac{3}{5}} = \frac{\frac{1}{3}}{1-\frac{2}{5}} = \frac{\frac{1}{3}}{\frac{3}{5}} = \frac{1}{3}\cdot \frac{5}{3} = \frac{5}{9}.\] Therefore, $P = \frac{5}{9}$, so the answer is $9-5 = \boxed{\textbf{(D)}\ 4}.$

Solution by ktong

~minor LaTeX edit by virjoy2001

Video Solution

https://youtu.be/IRyWOZQMTV8?t=4552

~ pi_is_3.14

Video Solution

https://www.youtube.com/watch?v=umr2Aj9ViOA

See Also

2017 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 17
Followed by
Problem 19
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS