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

m (Solution)
m (See also)
 
(7 intermediate revisions by 2 users not shown)
Line 3: Line 3:
  
 
== 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=\boxed{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 ==

Latest revision as of 00:07, 25 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 (ProblemsAnswer KeyResources)
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