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IN a tournament with 5 teams,each team plays every other team once.Each game ends in a win for one of the two teams.Each team has 1/2 chance of winning each game.The probability that no team wins all it's game or loses all it's games is:

I can't understand how to apply binomial distribution in this question.I always apply that method when I can visualise the sample space and know which outcomes I want but I can't seem to do that here.Can somebody help me .

P.S-The answer given is 17/32

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  • $\begingroup$ Duplicate $\endgroup$ – Remy Nov 19 '17 at 6:48
  • $\begingroup$ I want it to be solved by binomial distribution method not be permutation and see where i lack in in understanding the formulae $\endgroup$ – Hydrous Caperilla Nov 19 '17 at 6:55
  • $\begingroup$ There is no way I can think of to solve this without using the inclusion-exclusion principle, but I tried to solve it in a way that most relates to the binomial distribution. $\endgroup$ – Remy Nov 19 '17 at 7:17
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Each team plays $4$ games.

$P(\text{a team wins all its games}) = {5\choose{1}}\cdot{1\over{2}}^4$

$P(\text{a team loses all its games}) = {5\choose{1}}\cdot{1\over{2}}^4$

$P(\text{a team wins all its games and any remaining team loses all its games}) = {5\choose{1}} \cdot {4\choose{1}} \cdot {1\over2}^7$

$P(\text{at least one team wins or loses all their games}) = {5\choose{1}}\cdot{1\over{2}}^4+{5\choose{1}}\cdot{1\over{2}}^4-{5\choose{1}} \cdot {4\choose{1}} \cdot {1\over2}^7={15\over{32}}$

$$\begin{align*} P(\text{no team wins or loses all its games}) &= 1 - P(\text{at least one team wins or loses all their games})\\\\ &= 1 - \frac{15}{32} \\\\ &= \frac{17}{32} \\\\ \end{align*}$$

Perhaps the most tricky calculation here was

$P(\text{a team wins all its games and any remaining team loses all its games}) = {5\choose{1}} \cdot {4\choose{1}} \cdot {1\over2}^7$

This comes from choosing one team of the $5$ to win all its games with probability $\frac{1}{2}^4$ and then choosing one of the remaining $4$ teams to lose all it's games with probability $\frac{1}{2}^3$ since we already know they lose to the team that won all its games.

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