Management Aptitude Test Sep 2013 Paper

(b) Probability that A wins = $\frac{6}{12}=\frac{1}{2}$
Probability that B wins = $\frac{4}{12}=\frac{1}{3}$
Probability that game is drawn = $\frac{1}{6}$
Now, the probability that A wins 1st game in series of 3 games
$\frac{1}{2}\times \frac{1}{3}\times \frac{1}{2}=\frac{1}{12}$
Also, the probability that B wins in 1st in series of 3 games
$\frac{1}{3}\times \frac{1}{2}\times \frac{1}{3}=\frac{1}{18}$
Now, probability that they win alternately
= $\frac{1}{12}+\frac{1}{18}$ = $\frac{3+2}{36}=\frac{5}{36}$