Difference between revisions of "AoPS Wiki talk:Problem of the Day/July 15, 2011"
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(Corrected arithmetic error: C(10,6) = 210, not 270.) |
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==Solution== | ==Solution== | ||
− | 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 | + | 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{210+120+45+10+1}{1024}=\frac{386}{1024}=\boxed{\frac{193}{512}}</math>. |
Latest revision as of 12:29, 20 June 2012
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 .