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 points 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. $\frac{1}{4}$
• B. $\frac{5}{12}$
• C. $\frac{1}{2}$
• D. $\frac{7}{12}$

Answer 1

The correct option is B

$\frac{5}{12}$

Here, $P(X>Y)=P\left(T_1\text{win}\right)P\left(T_1\text{win}\right)+P\left(T_1\text{win}\right)P\left(\text{draw}\right)+P\left(\text{draw}\right)P\left(T_1\text{win}\right)=(\frac{1}{2}\times \frac{1}{2})+(\frac{1}{2}\times \frac{1}{6})+(\frac{1}{6}\times \frac{1}{2})=\frac{5}{12}$