Probability of Team T1 Outscoring Team T2 in Two Matches

To find the probability that $X > Y$, where $X$ and $Y$ represent the total points accumulated by teams $T_1$ and $T_2$ respectively after two games, we first define the outcomes for each game for team $T_1$:

• Win ($W$): $T_1$ gets 3 points, $T_2$ gets 0 points. $P(W) = \frac{1}{2}$
• Draw ($D$): $T_1$ gets 1 point, $T_2$ gets 1 point. $P(D) = \frac{1}{6}$
• Loss ($L$): $T_1$ gets 0 points, $T_2$ gets 3 points. $P(L) = \frac{1}{3}$

Since the two matches are independent, there are $3 \times 3 = 9$ possible outcomes for the pair of matches. Let us evaluate the points $(X, Y)$ for each possible outcome:

1. Outcome $WW$: $X = 3 + 3 = 6, \quad Y = 0 + 0 = 0 \implies X > Y$ $P(WW) = P(W) \cdot P(W) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$

2. Outcome $WD$: $X = 3 + 1 = 4, \quad Y = 0 + 1 = 1 \implies X > Y$ $P(WD) = P(W) \cdot P(D) = \frac{1}{2} \times \frac{1}{6} = \frac{1}{12}$

3. Outcome $WL$: $X = 3 + 0 = 3, \quad Y = 0 + 3 = 3 \implies X = Y$

4. Outcome $DW$: $X = 1 + 3 = 4, \quad Y = 1 + 0 = 1 \implies X > Y$ $P(DW) = P(D) \cdot P(W) = \frac{1}{6} \times \frac{1}{2} = \frac{1}{12}$

5. Outcome $DD$: $X = 1 + 1 = 2, \quad Y = 1 + 1 = 2 \implies X = Y$

6. Outcome $DL$: $X = 1 + 0 = 1, \quad Y = 1 + 3 = 4 \implies X < Y$

7. Outcome $LW$: $X = 0 + 3 = 3, \quad Y = 3 + 0 = 3 \implies X = Y$

8. Outcome $LD$: $X = 0 + 1 = 1, \quad Y = 3 + 1 = 4 \implies X < Y$

9. Outcome $LL$: $X = 0 + 0 = 0, \quad Y = 3 + 3 = 6 \implies X < Y$

The condition $X > Y$ is satisfied for the mutually exclusive outcomes $\{WW, WD, DW\}$.

Therefore, the required probability $P(X > Y)$ is given by: $P(X > Y) = P(WW) + P(WD) + P(DW)$ $P(X > Y) = \frac{1}{4} + \frac{1}{12} + \frac{1}{12}$ $P(X > Y) = \frac{1}{4} + \frac{2}{12} = \frac{1}{4} + \frac{1}{6} = \frac{3 + 2}{12} = \frac{5}{12}$

Thus, the correct option is B.