5 teams compete against each other only once the probability that a team wins all four of its games and the other team loses all 4 of its games

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- Apr 2nd 2012, 02:30 AMprasumProbability
5 teams compete against each other only once the probability that a team wins all four of its games and the other team loses all 4 of its games

- Apr 2nd 2012, 02:43 AMVinceeeRe: Probability
Not enough info given?

- Apr 3rd 2012, 01:07 PMMathoManRe: Probability
Think in terms of 'league table' listing the number of wins, draws, and losses for each team. Be aware that each one of 5 teams has played 4 games since each team has four opposing teams.

One such possible 'league table' could look like:

W D L

4 0 0 --> 4 wins

1 2 1

1 2 1

1 2 1

0 0 4 --> 4 losses

Take it from there. - Apr 4th 2012, 02:29 AMprasumRe: Probability
Can you tell me the answer of this question

- Apr 4th 2012, 03:01 AMprincepsRe: Probability
All possible outcomes for one team are given in following table :

$\displaystyle \begin{array}{|c|c|c|} W & D & L \\\hline4&0&0\\3&1&0\\3&0&1\\2&1&1\\2&0&2\\2&2&0\\ 1&0&3\\1&3&0\\1&1&2\\1&2&1\\0&1&3\\0&3&1\\0&2&2\\0 &4&0\\0&0&4\\\end{array}$

Hence , probability for both cases is :

$\displaystyle P=\frac{1}{15}$ - Apr 4th 2012, 03:11 AMa tutorRe: Probability
Some big assumptions there.

:eek: