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

Revision as of 13:10, 21 January 2021

Problem

A coin that comes up heads with probability $p > 0$ and tails with probability $1 - p > 0$ independently on each flip is flipped $8$ times. Suppose that 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 $5(1-p)=p$ . We can do this because we know that both $p$ and $1-p$ are between $0$ and $1$ , so they are both positive. Now, we have:

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

Now, we get $5+6=\boxed{011}$ .

~Jerry_Guo

Solution 3

Rewrite it as : $(P)^3$ $(1-P)^5$ = $\frac {1}{25}<cmath>(P)^5</cmath>(P)^3$

This can be simplified as $24P^2 -50P + 25 = 0$

The can be factored into: $(4P-5)(6P-5)$

This yields two solutions: $5/4$ (ignored) and $\boxed {5/6}$

Video Solution

https://youtu.be/NL79UexadzE

~IceMatrix

Video Solution 2

https://www.youtube.com/watch?v=P00iOJdQiL4

~Shreyas S

@@ Line 29: / Line 29: @@
 ~Jerry_Guo
+== Solution 3 ==
+Rewrite it as : <math>(P)^3</math><math>(1-P)^5</math>=<math>\frac {1}{25}<cmath>(P)^5</cmath>(P)^3</math>
+This can be simplified as <math>24P^2 -50P + 25 = 0</math>
+The can be factored into: <math>(4P-5)(6P-5)</math>
+This yields two solutions: <math>5/4</math>(ignored) and <math> \boxed {5/6}</math>
 ==Video Solution==

2009 AIME I (Problems • Answer Key • Resources)
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