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

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== Solution ==
 
== Solution ==
{{solution}}
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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(\frac{1}{3})^3(\frac{2}{3})^2 = \frac{40}{243}</math>, so <math>i+j=40+243 = \mathrm{283}</math>.
  
 
== See also ==
 
== See also ==
* [[1989 AIME Problems/Problem 6|Next Problem]]
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{{AIME box|year=1989|num-b=4|num-a=6}}
* [[1989 AIME Problems/Problem 4|Previous Problem]]
 
* [[1989 AIME Problems]]
 

Revision as of 21:29, 26 February 2007

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

Denote the probability of getting a heads in one flip of the biased coins 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 ${5\choose3}(h)^3(1-h)^2 = 10(\frac{1}{3})^3(\frac{2}{3})^2 = \frac{40}{243}$, so $i+j=40+243 = \mathrm{283}$.

See also

1989 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
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All AIME Problems and Solutions
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