15) Four basketball teams of equal strength compete in a round-robin tournament (each team plays every other team once). What is the probability that there is a team that has won all of its games at the end of the tournament? (basketball games cannot end in a tie).
(A) $\frac{1}{6}$
(B) $\frac{1}{\pi}$
(C) $\frac{1}{3}$
(D) $\frac{1}{2}$
(E) $\frac{2}{3}$.
15) The answer is $\mathbf{( D )}$.

The probability that a single team wins all of its matches (which are three) is clearly $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2}=\frac{1}{8}$.

Since there are four teams, the probability that one of them wins all of its matches (it being impossible for two teams to win all of their matches) is $4 \cdot \frac{1}{8}=\frac{1}{2}$.

## Second Solution

After the first match day, two teams have won their respective matches. In the subsequent match between these two teams, one of them wins. This is the only team that can win all of its matches, and this happens only if it wins its third match, and thus with a probability of $\frac{1}{2}$.