Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1989 AIME Problems/Problem 5"

m (→‎Solution: l,r ())
m (→‎See also)
 
(11 intermediate revisions by 6 users not shown)
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
−
When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to <math>0^{}_{}</math> and is the same as that of getting heads exactly twice. Let <math>\frac ij^{}_{}</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3_{}^{}</math> out of <math>5^{}_{}</math> flips. Find <math>i+j^{}_{}</math>.
+
When a certain biased coin is flipped five times, the [[probability]] of getting heads exactly once is not equal to <math>0</math> and is the same as that of getting heads exactly twice. Let <math>\frac ij</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3</math> out of <math>5</math> flips. Find <math>i+j</math>.
  
 
== Solution ==
 
== Solution ==
−
Denote the probability of getting a heads in one flip of the biased coins as <math>h</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3</math>. After canceling out terms, we get <math>1 - h = 2h</math>, so <math>h = \frac{1}{3}</math>. The answer we are looking for is <math>{5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}</math>, so <math>i+j=40+243 = \mathrm{283}</math>.
+
===Solution 1 ===
 +
Denote the probability of getting a heads in one flip of the biased coin as <math>h</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3</math>. After canceling out terms, we get <math>1 - h = 2h</math>, so <math>h = \frac{1}{3}</math>. The answer we are looking for is <math>{5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}</math>, so <math>i+j=40+243=\boxed{283}</math>.
 +
 
 +
 
 +
=== Solution 2 ===
 +
Denote the probability of getting a heads in one flip of the biased coins as <math>h</math> and the probability of getting a tails as <math>t</math>. Based upon the problem, note that <math>{5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3</math>. After cancelling out terms, we end up with <math>t = 2h</math>. To find the probability getting <math>3</math> heads, we need to find <math>{5\choose3}\dfrac{(h)^3(t)^2}{(h + t)^5} =10\cdot\dfrac{(h)^3(2h)^2}{(h + 2h)^5}</math> (recall that <math>h</math> cannot be <math>0</math>). The result after simplifying is <math>\frac{40}{243}</math>, so <math>i + j = 40 + 243 = \boxed{283}</math>.
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=1989|num-b=4|num-a=6}}
 
{{AIME box|year=1989|num-b=4|num-a=6}}
 +
 +
[[Category:Intermediate Combinatorics Problems]]
 +
{{MAA Notice}}

Latest revision as of 23:07, 24 June 2023

Problem

When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to $0$ and is the same as that of getting heads exactly twice. Let $\frac ij$, in lowest terms, be the probability that the coin comes up heads in exactly $3$ out of $5$ flips. Find $i+j$.

Solution

Solution 1

Denote the probability of getting a heads in one flip of the biased coin as $h$. Based upon the problem, note that ${5\choose1}(h)^1(1-h)^4 = {5\choose2}(h)^2(1-h)^3$. After canceling out terms, we get $1 - h = 2h$, so $h = \frac{1}{3}$. The answer we are looking for is ${5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}$, so $i+j=40+243=\boxed{283}$.


Solution 2

Denote the probability of getting a heads in one flip of the biased coins as $h$ and the probability of getting a tails as $t$. Based upon the problem, note that ${5\choose1}(h)^1(t)^4 = {5\choose2}(h)^2(t)^3$. After cancelling out terms, we end up with $t = 2h$. To find the probability getting $3$ heads, we need to find ${5\choose3}\dfrac{(h)^3(t)^2}{(h + t)^5} =10\cdot\dfrac{(h)^3(2h)^2}{(h + 2h)^5}$ (recall that $h$ cannot be $0$). The result after simplifying is $\frac{40}{243}$, so $i + j = 40 + 243 = \boxed{283}$.

See also

1989 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1989_AIME_Problems/Problem_5&oldid=194765"