Difference between revisions of "2009 AIME I Problems/Problem 3"
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A coin that comes up heads with probability <math>p > 0</math> and tails with probability <math>1 - p > 0</math> independently on each flip is flipped <math>8</math> times. Suppose that the probability of three heads and five tails is equal to <math>\frac {1}{25}</math> of the probability of five heads and three tails. Let <math>p = \frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | A coin that comes up heads with probability <math>p > 0</math> and tails with probability <math>1 - p > 0</math> independently on each flip is flipped <math>8</math> times. Suppose that the probability of three heads and five tails is equal to <math>\frac {1}{25}</math> of the probability of five heads and three tails. Let <math>p = \frac {m}{n}</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m + n</math>. | ||
− | == Solution == | + | == Solution 1 == |
The probability of three heads and five tails is <math>\binom {8}{3}p^3(1-p)^5</math> and the probability of five heads and three tails is <math>\binom {8}{3}p^5(1-p)^3</math>. | The probability of three heads and five tails is <math>\binom {8}{3}p^3(1-p)^5</math> and the probability of five heads and three tails is <math>\binom {8}{3}p^5(1-p)^3</math>. | ||
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== Solution 2 == | == Solution 2 == | ||
− | We start as shown above. However, | + | We start as shown above. However, when we get to <math>25(1-p)^2=p^2</math>, we square root both sides to get <math>5(1-p)=p</math>. We can do this because we know that both <math>p</math> and <math>1-p</math> are between <math>0</math> and <math>1</math>, so they are both positive. Now, we have: |
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
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~Jerry_Guo | ~Jerry_Guo | ||
+ | |||
+ | == Solution 3 == | ||
+ | Rewrite it as : <math>(P)^3</math><math>(1-P)^5=\frac {1}{25}</math> <math>(P)^5</math><math>(1-P)^3</math> | ||
+ | |||
+ | This can be simplified as <math>24P^2 -50P + 25 = 0</math> | ||
+ | |||
+ | This can be factored into <math>(4P-5)(6P-5)</math> | ||
+ | |||
+ | This yields two solutions: <math>\frac54</math> (ignored because it would result in <math>1-p<0</math> ) or <math>\frac56</math> | ||
+ | |||
+ | Therefore, the answer is <math>5+6</math> = <math>\boxed {011}</math> | ||
==Video Solution== | ==Video Solution== |
Latest revision as of 03:35, 16 January 2022
Contents
Problem
A coin that comes up heads with probability and tails with probability independently on each flip is flipped times. Suppose that the probability of three heads and five tails is equal to of the probability of five heads and three tails. Let , where and are relatively prime positive integers. Find .
Solution 1
The probability of three heads and five tails is and the probability of five heads and three tails is .
Therefore, the answer is .
Solution 2
We start as shown above. However, when we get to , we square root both sides to get . We can do this because we know that both and are between and , so they are both positive. Now, we have:
Now, we get .
~Jerry_Guo
Solution 3
Rewrite it as :
This can be simplified as
This can be factored into
This yields two solutions: (ignored because it would result in ) or
Therefore, the answer is =
Video Solution
~IceMatrix
Video Solution 2
https://www.youtube.com/watch?v=P00iOJdQiL4
~Shreyas S
See also
2009 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
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.