Difference between revisions of "2009 AIME I Problems/Problem 3"
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== Solution == | == Solution == | ||
| − | + | 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>. | |
| − | |||
| − | <cmath>25 | + | <cmath>\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 \ | |
| − | + | 5(1-p)&=p \ | |
| − | + | 5-5p&=p \ | |
| − | + | 5&=6p \ | |
| − | + | p&=\frac {5}{6}\end{align*}</cmath> | |
| − | < | + | |
| + | Therefore, the answer is <math>5+6=\boxed{11}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2009|n=I|num-b=2|num-a=4}} | {{AIME box|year=2009|n=I|num-b=2|num-a=4}} | ||
Revision as of 17:36, 26 March 2009
Problem
A coin that comes up heads with probability 
and tails with probability 
independently on each flip is flipped eight times. Suppose 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
The probability of three heads and five tails is 
and the probability of five heads and three tails is 
.
Therefore, the answer is 
.
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 | ||
