Difference between revisions of "2009 AIME I Problems/Problem 3"

Revision as of 15:51, 2 June 2020

Problem

A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped eight times. Suppose the probability of three heads and five tails is equal to $\frac {1}{25}$ of the probability of five heads and three tails. Let $p = \frac {m}{n}$ , where $m$ and $n$ are relatively prime positive integers. Find $m + n$ .

Solution

The probability of three heads and five tails is $\binom {8}{3}p^3(1-p)^5$ and the probability of five heads and three tails is $\binom {8}{3}p^5(1-p)^3$ .

$\begin{align*} 25\binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\ 25(1-p)^2&=p^2 \\ 25p^2-50p+25&=p^2 \\ 24p^2-50p+25&=0 \\ p&=\frac {5}{6}\end{align*}$

Therefore, the answer is $5+6=\boxed{011}$ .

Solution 2

We start as shown above, however, as we get to $25(1-p)^2=p^2$ , we square root both sides to get $5(1-p)=p$ . Then, we get:

$\begin{align*} 5(1-p)&=p \\ 5-5p&=p \\ 5&=6p \\ p&=\frac {5}{6}\end{align*}$

We get the same answer as above, $5+6=\boxed{011}$

-Jerry_Guo

Video Solution

https://youtu.be/NL79UexadzE

~IceMatrix

@@ Line 10: / Line 10: @@
 \binom {8}{3}p^3(1-p)^5&=\binom {8}{3}p^5(1-p)^3 \\
 (1-p)^2&=p^2 \\
-p^2-50p+25=p^2 \\
+p^2-50p+25&=p^2 \\
-p^2-50p+25=0 \\
+p^2-50p+25&=0 \\
 p&=\frac {5}{6}\end{align*}</cmath>
 Therefore, the answer is <math>5+6=\boxed{011}</math>.
+== Solution 2 ==
+We start as shown above, however, as we get to <math>25(1-p)^2=p^2</math>, we square root both sides to get <math>5(1-p)=p</math>. Then, we get:
+<cmath>\begin{align*}
+(1-p)&=p \\
+-5p&=p \\
+&=6p \\
+p&=\frac {5}{6}\end{align*}</cmath>
+We get the same answer as above, <math>5+6=\boxed{011}</math>
+-Jerry_Guo
 ==Video Solution==

2009 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 2	Followed by Problem 4
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