1996 AIME Problems/Problem 6

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 product 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 (ProblemsAnswer KeyResources)
Preceded by
Problem 5
Followed by
Problem 7
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All AIME Problems and Solutions

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