Question

In a game of chess, the chance of $A$'s win is three times the chance of a draw, and the chance of $B$'s win is two times the chance of a draw. If they agree to play 3 games, then the probability that $A$ and $B$ win alternatively, is

• A. $\frac{1}{8}$
• B. $\frac{5}{72}$
• C. $\frac{7}{8}$
• D. $\frac{5}{36}$

Answer 1

The correct option is D $\frac{5}{36}$

Suppose the probability of a draw is $P\left(D\right)=a$
Hence, probability of A's win$=3a$
Also, the probability of B's win$=2a$
$\therefore 3a+2a+a=1\Rightarrow a=\frac{1}{6}$ $(\because$ the events are exhaustive and mutually exclusive $)$
$\Rightarrow P\left(A\right)=3a=\frac{1}{2},P\left(B\right)=2a=\frac{1}{3}$
In the three games played, if A and B win alternatively, two sequences of win are possible: $ABA$ or $BAB$.
The probability of the occurrence of the first case is $\frac{1}{2}\times \frac{1}{3}\times \frac{1}{2}=\frac{1}{12}.$
The probability of the occurrence of the second case is $\frac{1}{3}\times \frac{1}{2}\times \frac{1}{3}=\frac{1}{18}.$
Hence, the total probability of alternate win is $\frac{1}{12}+\frac{1}{18}=\frac{5}{36}$.