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 `(1)/(2),(1)/(6) and (1)/(3)` respectively. Each team gets 3 points for a win. 1 point for a draw and 10 pont 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.

Options

A. `(1)/(4)`
B. `(5)/(12)`
C. `(1)/(2)`
D. `(7)/(12)`

Answer

To solve the problem, we need to analyze the outcomes of the two games played between teams \( T_1 \) and \( T_2 \) and calculate the total points scored by each team. We will then find the probability that the total points scored by \( T_1 \) equals the total points scored by \( T_2 \).

### Step 1: Define the outcomes and their probabilities
The outcomes of a single game between \( T_1 \) and \( T_2 \) are:
- \( T_1 \) wins: Probability \( P(W) = \frac{1}{2} \)
- Draw: Probability \( P(D) = \frac{1}{6} \)
- \( T_2 \) wins: Probability \( P(L) = \frac{1}{3} \)

### Step 2: Points awarded for each outcome
Points awarded for each outcome are:
- Win: 3 points
- Draw: 1 point
- Loss: 10 points

### Step 3: Calculate total points for each possible outcome after two games
We need to consider the different combinations of outcomes for two games. The possible outcomes are:
1. \( W, W \) (T1 wins both)
2. \( W, D \) (T1 wins one, draws one)
3. \( D, W \) (T1 draws one, wins one)
4. \( D, D \) (Both teams draw)
5. \( L, W \) (T1 loses one, wins one)
6. \( W, L \) (T1 wins one, loses one)
7. \( L, D \) (T1 loses one, draws one)
8. \( D, L \) (T1 draws one, loses one)
9. \( L, L \) (T1 loses both)

Now we calculate the total points \( X \) for \( T_1 \) and \( Y \) for \( T_2 \) for each outcome:

1. \( W, W \): \( X = 3 + 3 = 6 \), \( Y = 10 + 10 = 20 \)
2. \( W, D \): \( X = 3 + 1 = 4 \), \( Y = 10 + 3 = 13 \)
3. \( D, W \): \( X = 1 + 3 = 4 \), \( Y = 3 + 10 = 13 \)
4. \( D, D \): \( X = 1 + 1 = 2 \), \( Y = 3 + 3 = 6 \)
5. \( L, W \): \( X = 10 + 3 = 13 \), \( Y = 3 + 10 = 13 \)
6. \( W, L \): \( X = 3 + 10 = 13 \), \( Y = 10 + 3 = 13 \)
7. \( L, D \): \( X = 10 + 1 = 11 \), \( Y = 3 + 10 = 13 \)
8. \( D, L \): \( X = 1 + 10 = 11 \), \( Y = 10 + 3 = 13 \)
9. \( L, L \): \( X = 10 + 10 = 20 \), \( Y = 3 + 3 = 6 \)

### Step 4: Identify outcomes where \( X = Y \)
From the outcomes calculated, we see that \( X = Y \) occurs in the following cases:
- Case 2: \( W, D \) (4, 13)
- Case 3: \( D, W \) (4, 13)
- Case 5: \( L, W \) (13, 13)
- Case 6: \( W, L \) (13, 13)

### Step 5: Calculate the probabilities of these outcomes
Now we calculate the probabilities of the outcomes where \( X = Y \):

1. For \( W, W \): \( P(W) \cdot P(W) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \)
2. For \( W, D \): \( P(W) \cdot P(D) = \frac{1}{2} \cdot \frac{1}{6} = \frac{1}{12} \)
3. For \( D, W \): \( P(D) \cdot P(W) = \frac{1}{6} \cdot \frac{1}{2} = \frac{1}{12} \)
4. For \( D, D \): \( P(D) \cdot P(D) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} \)
5. For \( L, W \): \( P(L) \cdot P(W) = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6} \)
6. For \( W, L \): \( P(W) \cdot P(L) = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} \)

### Step 6: Sum the probabilities of outcomes where \( X = Y \)
The total probability \( P(X = Y) \) is the sum of the probabilities of the cases where \( X = Y \):
\[ P(X = Y) = P(L, W) + P(W, L) = \frac{1}{6} + \frac{1}{6} = \frac{1}{3} \]

### Final Answer
Thus, the probability that the total points scored by \( T_1 \) equals the total points scored by \( T_2 \) after two games is:
\[ \boxed{\frac{13}{36}} \]