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
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.
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 :
$\displaystyle P=\frac{1}{15}$