Difference between revisions of "1989 AIME Problems/Problem 5"
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== Solution == | == Solution == | ||
| − | 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\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}</math>, so <math>i+j=40+243 = \ | + | 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\left(\frac{1}{3}\right)^3\left(\frac{2}{3}\right)^2 = \frac{40}{243}</math>, so <math>i+j=40+243=\boxed{283}</math>. |
== See also == | == See also == | ||
Revision as of 20:31, 10 October 2016
Problem
When a certain biased coin is flipped five times, the probability of getting heads exactly once is not equal to 
and is the same as that of getting heads exactly twice. Let 
, in lowest terms, be the probability that the coin comes up heads in exactly 
out of 
flips. Find 
.
Solution
Denote the probability of getting a heads in one flip of the biased coins as 
. Based upon the problem, note that 
. After canceling out terms, we get 
, so 
. The answer we are looking for is 
, so 
.
See also
| 1989 AIME (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
