Football teams `T_(1)and T_(2)` have to play two games are independent. The probabilities of `T_(1)` winning, drawing and lossing a game against `T_(2)` are `1/6,1/6and 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

Options

A. `11/36`
B. `1/3`
C. `13/36`
D. `1/2`

Explanation

To solve the problem, we need to determine the total points scored by teams \( T_1 \) and \( T_2 \) after two games and find the probability that both teams score the same number of points, \( P(X = Y) \).

### Step 1: Understand the outcomes of a single game
The outcomes of a single game between \( T_1 \) and \( T_2 \) can be:
- \( T_1 \) wins (W)
- Draw (D)
- \( T_2 \) wins (L)

The probabilities for these outcomes are:
- \( P(W) = \frac{1}{6} \)
- \( P(D) = \frac{1}{6} \)
- \( P(L) = \frac{1}{3} \)

Since these are the only outcomes, we can find \( P(L) \) as follows:
\[ P(L) = 1 - P(W) - P(D) = 1 - \frac{1}{6} - \frac{1}{6} = \frac{2}{3} \]

### Step 2: Calculate the points for each outcome
The points awarded for each outcome are:
- If \( T_1 \) wins: \( T_1 \) gets 3 points, \( T_2 \) gets 0 points.
- If it's a draw: Both teams get 1 point each.
- If \( T_2 \) wins: \( T_1 \) gets 0 points, \( T_2 \) gets 3 points.

### Step 3: List the possible outcomes for two games
Since the games are independent, we can list the outcomes for two games:
1. \( WW \) (T1 wins both)
2. \( WD \) (T1 wins first, draws second)
3. \( WL \) (T1 wins first, T2 wins second)
4. \( DW \) (Draw first, T1 wins second)
5. \( DD \) (Draw both)
6. \( DL \) (Draw first, T2 wins second)
7. \( LW \) (T2 wins first, T1 wins second)
8. \( LD \) (T2 wins first, draws second)
9. \( LL \) (T2 wins both)

### Step 4: Calculate the points for each outcome
Now we calculate the points \( X \) and \( Y \) for each outcome:
1. \( WW \): \( X = 6, Y = 0 \)
2. \( WD \): \( X = 4, Y = 1 \)
3. \( WL \): \( X = 3, Y = 3 \)
4. \( DW \): \( X = 3, Y = 0 \)
5. \( DD \): \( X = 2, Y = 2 \)
6. \( DL \): \( X = 1, Y = 3 \)
7. \( LW \): \( X = 0, Y = 3 \)
8. \( LD \): \( X = 1, Y = 1 \)
9. \( LL \): \( X = 0, Y = 6 \)

### Step 5: Identify the outcomes where \( X = Y \)
The outcomes where \( X = Y \) are:
- \( WL \): \( X = 3, Y = 3 \)
- \( DD \): \( X = 2, Y = 2 \)
- \( LD \): \( X = 1, Y = 1 \)

### Step 6: Calculate the probabilities of these outcomes
Now we calculate the probabilities of these outcomes:
1. \( P(WL) = P(W) \cdot P(L) = \frac{1}{6} \cdot \frac{2}{3} = \frac{1}{9} \)
2. \( P(DD) = P(D) \cdot P(D) = \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} \)
3. \( P(LD) = P(L) \cdot P(D) = \frac{2}{3} \cdot \frac{1}{6} = \frac{1}{9} \)

### Step 7: Sum the probabilities
Now we sum the probabilities of the outcomes where \( X = Y \):
\[ P(X = Y) = P(WL) + P(DD) + P(LD) = \frac{1}{9} + \frac{1}{36} + \frac{1}{9} \]

To add these fractions, we need a common denominator:
\[ P(X = Y) = \frac{4}{36} + \frac{1}{36} + \frac{4}{36} = \frac{9}{36} = \frac{1}{4} \]

### Final Answer
Thus, the probability that both teams score the same number of points after two games is:
\[ P(X = Y) = \frac{1}{4} \]