Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 15, 2011"
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− | There are <math>\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}</math> ways to satisfy the baby storks, out of a total of <math>2^10</math> ways to catch the fish. Using the fact that <math>\binom{n}{k}=\binom{n}{n-k}</math>, we can quickly evaluate the probability as <math>\frac{270+120+45+10+1}{1024}=\frac{446}{1024}=\boxed{\frac{223}{512}}</math>. | + | There are <math>\binom{10}{6}+\binom{10}{7}+\binom{10}{8}+\binom{10}{9}+\binom{10}{10}</math> ways to satisfy the baby storks, out of a total of <math>2^{10}</math> ways to catch the fish. Using the fact that <math>\binom{n}{k}=\binom{n}{n-k}</math>, we can quickly evaluate the probability as <math>\frac{270+120+45+10+1}{1024}=\frac{446}{1024}=\boxed{\frac{223}{512}}</math>. |
Revision as of 04:52, 18 August 2011
Problem
AoPSWiki:Problem of the Day/July 15, 2011
Solution
There are ways to satisfy the baby storks, out of a total of ways to catch the fish. Using the fact that , we can quickly evaluate the probability as .