1996 AIME Problems/Problem 6

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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 $\frac{17}{32}$ .

$17+32=\boxed{049}$

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

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1996 AIME Problems/Problem 6

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Problem

Solution

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