BdMO Online Forum

there will be total $^5C_2=10$ matches and there can be $2^{10}$ results. let's count in how many cases there will be an undefeated team and/or winless team.

there can't be more than one undefeated team or winless team in a single case. so, number of cases in which one team wins all matches is $^5C_1\cdot2^6$. so is for the cases in which one team loses all.

so one team wins all or one team loses all in $^5C_1\cdot2^6+^5C_1\cdot2^6$ cases.

but we counted the cases twice in which there's an undefeated team AND a winless team. number of such cases are $^5P_2\cdot2^3$.

so probability$=\dfrac{2^{10}-(2\cdot^5C_1\cdot2^6-^5P_2\cdot2^3)}{2^{10}}$