Question

Football teams $T_1$ and $T_2$ have to play two games against each other. It is assumed that the outcomes of the two games are independent. The probabilities of $T_1$ winning, drawing and losing a game against $T_2$ are $\frac{1}{2},\frac{1}{6}$ and $\frac{1}{3},$ respectively. Each team gets $3$ points for a win, $1$ point for a draw and $0$ point for a loss in a game. Let $X$ and $Y$ denote the total points scored by teams $T_1$ and $T_2,$ respectively, after two games.

$P(X=Y)$ is

• A. $\frac{11}{36}$
• B. $\frac{1}{3}$
• C. $\frac{13}{36}$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $\frac{13}{36}$

$P(X=Y)=P\left(T_1\text{and}{T}_2\text{win alternately}\right)+P\left(\text{Both matches are draw}\right)$
$\Rightarrow P(X=Y)=(\frac{1}{2}\times \frac{1}{3}+\frac{1}{3}\times \frac{1}{2})+(\frac{1}{6}\times \frac{1}{6})$
$\therefore P(X=Y)=\frac{1}{3}+\frac{1}{36}=\frac{13}{36}$