Question

eight teams compete in a tournament. each pair of teams plays exactly one game against each other. there are no ties. if the two possible outcomes of each game are equally likely, what is the probability that every team loses at least one game and wins at least one game?

Answer 1

The probability that every team loses at least one game and wins at least one game is $\frac{903}{1024}$.

Let’s label the 8 teams as A, B, C, D, E, F, J, H.

First, we determine the total number of games played.

Because of every pair of teams plays exactly one game, so every team plays 7 games (one against each of the other 7 teams). And there are 8 teams, so it seems as if there are 8 x 7 = 56 games, except that every game has been counted twice in this total. So, there are in fact $\frac{8.7}{2}$= 28 games played.

Because of there are 28 games played and there are 2 equally likely outcomes for every game, so there are $2^{28}$ possible combinations of outcomes.

To find out the probability that every team loses at least one game and every team wins at least one game, we determine the probability that there is a team that loses 0 games or a team that wins 0 games and subtract this probability from 1.

And because we know that the total number of possible combinations of outcomes, we determine the probability by counting the number of combinations of outcomes in where there is a team that loses 0 games or a team that wins 0 games, or both.

To find out the number of combinations of outcomes in where there is a team which wins all of its games, we mark that there are 8 ways to choose this team. Once a team is chosen (we call this team X), the results of the 7 games played by X are determined (X wins all of these) and the outcomes of the remaining 28 − 7 = 21 games are undetermined.

And because of there are two possible outcomes for each of these 21 undetermined games, so there are 8 x $2^{21}$ combinations of outcomes in which there is a team that wins all of its games. Similarly, there are 8 x $2^{21}$ combinations of outcomes in which there is a team that loses all of its games.

Now, we note that there might be combinations of outcomes that are involved in both of these counts. There might be combinations of outcomes in where there is a team that wins all of its games and in where there is a team that loses all of its games.

Since this total has been included in both sets of 8 x $2^{21}$ combinations of outcomes, we have to determine this total and subtract it once to leave these combinations included exactly once in our total.

To determine the number of combinations of outcomes in this case, we choose a team (X) to win all of its games and a team (Y) to lose all of its games.

Once X is selected, the outcomes of its 7 games are all found (X wins).

Once Y is selected, the outcomes of its 6 additional games are all found (Y loses these 6 games plus the game with X that has already been found).

The outcomes of the remaining 28 − 7 − 6 = 15 games are undetermined.

Therefore, the number of combinations of outcomes is 8 x 7 x $2^{15}$ since there are 8 ways of choosing X, and then 7 ways of choosing Y (any team but X), and then $2^{15}$ combinations of outcomes for the undetermined games.

So, there are 8 x $2^{21}$ + 8 x $2^{21}$ − 8 x 7 x $2^{15}$ combinations of outcomes in where either one team loses 0 games or one team wins 0 games (or both).

Accordingly, the probability that one team loses 0 games or one team wins 0 games is

$\frac{8.2^{21} + 8.2^{21} - 8.7.2^{15} }{2^{28} }$ = $\frac{2^{15} (8.2^{6} + 8.2^{6} - 8.7)}{2^{28} }$ = $\frac{2^{3} . 2^{6} + 2^{3} . 2^{6} - 2^{3} . 7}{2^{13}}$ = $\frac{2^{6} + 2^{6} - 7}{2^{10}}$

It means that the probability that every team loses at least one game and wins at least one game is 1 − $\frac{64+64-7}{1024}$ = 1 − $\frac{121}{1024}$ = $\frac{903}{1024}$.

Learn more about probability at: https://brainly.com/question/11234923

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