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

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== Problem ==
 
== Problem ==
When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to <math>0^{}_{}</math> and is the same as that of getting heads exactly twice. Let <math>\frac ij^{}_{}</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3_{}^{}</math> out of <math>5^{}_{}</math> flips. Find <math>i+j^{}_{}</math>.
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When a certain biased coin is flipped five times, the [[probability]] of getting heads exactly once is not equal to <math>0^{}_{}</math> and is the same as that of getting heads exactly twice. Let <math>\frac ij^{}_{}</math>, in lowest terms, be the probability that the coin comes up heads in exactly <math>3_{}^{}</math> out of <math>5^{}_{}</math> flips. Find <math>i+j^{}_{}</math>.
  
 
== Solution ==
 
== Solution ==
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== See also ==
 
== See also ==
 
{{AIME box|year=1989|num-b=4|num-a=6}}
 
{{AIME box|year=1989|num-b=4|num-a=6}}
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[[Category:Intermediate Combinatorics Problems]]

Revision as of 18:51, 23 October 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 ${5\choose3}(h)^3(1-h)^2 = 10\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}$, so $i+j=40+243 = \mathrm{283}$.

See also

1989 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 4
Followed by
Problem 6
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions
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