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1996 AIME Problems/Problem 6: Difference between revisions

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== Problem ==
== Problem ==
In a five-team tournament, each team plays one game with every other team. Each team has a <math>50\%</math> chance of winning any game it plays. (There are no ties.) Let <math>\dfrac{m}{n}</math> be the probability that the tournament will product neither an undefeated team nor a winless team, where <math>m</math> and <math>n</math> are relatively prime integers. Find <math>m+n</math>.
In a five-team tournament, each team plays one game with every other team. Each team has a <math>50\%</math> chance of winning any game it plays. (There are no ties.) Let <math>\dfrac{m}{n}</math> be the probability that the tournament will produce neither an undefeated team nor a winless team, where <math>m</math> and <math>n</math> are relatively prime integers. Find <math>m+n</math>.


== Solution ==
== Solution ==

Revision as of 14:22, 17 February 2018

Problem

In a five-team tournament, each team plays one game with every other team. Each team has a $50\%$ chance of winning any game it plays. (There are no ties.) Let $\dfrac{m}{n}$ be the probability that the tournament will produce neither an undefeated team nor a winless team, where $m$ and $n$ are relatively prime integers. Find $m+n$.

Solution

We can use complementary counting: finding the probability that at least one team wins all games or at least one team loses all games.

No more than 1 team can win or lose all games, so at most one team can win all games and at most one team can lose all games.

Now we use PIE:

The probability that one team wins all games is $5\cdot (\frac{1}{2})^4=\frac{5}{16}$.

Similarity, the probability that one team loses all games is $\frac{5}{16}$.

The probability that one team wins all games and another team loses all games is $(5\cdot (\frac{1}{2})^4)(4\cdot (\frac{1}{2})^3)=\frac{5}{32}$.

$\frac{5}{16}+\frac{5}{16}-\frac{5}{32}=\frac{15}{32}$

Since this is the opposite of the probability we want, we subtract that from 1 to get $\frac{17}{32}$.

$17+32=\boxed{049}$

See also

1996 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 5 		Followed by
Problem 7
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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