Understanding this probability 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
 A: 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.
