14.19 $A, B$ two teams are in a series of matches. If either team has an equal chance of winning

a game, but to win the series, team $A$ must win 2 games, and team $B$ must win 3 games. Then the probability that team $A$ wins is
(A) 11 to 5.
(B) 5 to 2.
(C) 8 to 3.
(D) 3 to 2.
(E) 13 to 6.
(22nd American High School Mathematics Examination, 1971)
[Solution] Since the two teams will play 5 matches, the total number of different win-loss scenarios for each match is $\frac{2^{5}}{2}=$
16. In the matches, the four different winning sequences for team $A$ to lose are
$$
B B B\left(\frac{1}{8}\right), \quad A B B B\left(\frac{1}{16}\right), \quad B A B B\left(\frac{1}{16}\right), \quad B B A B\left(\frac{1}{16}\right) .
$$
(The numbers in parentheses after each sequence represent their probabilities). The total probability of team $A$ losing is $\frac{5}{16}$.
Therefore, the probability of winning is $\frac{11}{16}$.
Thus, the likelihood of team $A$ winning is 11 to 5.
Hence, the answer is $(A)$.