If tow loaded dice each have the property that 2 or 4 is three times as likely to appear as 1, 3, 5, or 6 on each roll. When two such dice are rolled, the probability of obtaining a total of 7 is `p ,` then the value of`[1//p]` is, where `[x]` represents the greatest integer less than or equal to `xdot`

Explanation

To solve the problem, we need to find the probability of obtaining a total of 7 when rolling two loaded dice, where the probabilities of rolling certain numbers are not uniform. Let's break down the solution step by step.

### Step 1: Define the probabilities
Let the probability of rolling a 1, 3, 5, or 6 be \( x \). According to the problem, the probabilities for rolling a 2 or 4 are three times as likely as rolling a 1, 3, 5, or 6. Therefore, we can express the probabilities as follows:
- \( P(1) = x \)
- \( P(2) = 3x \)
- \( P(3) = x \)
- \( P(4) = 3x \)
- \( P(5) = x \)
- \( P(6) = x \)

### Step 2: Set up the equation for total probability
The sum of all probabilities must equal 1:
\[ P(1) + P(2) + P(3) + P(4) + P(5) + P(6) = 1 \]
Substituting the probabilities we defined:
\[ x + 3x + x + 3x + x + x = 1 \]
This simplifies to:
\[ 10x = 1 \]

### Step 3: Solve for \( x \)
From the equation \( 10x = 1 \), we can solve for \( x \):
\[ x = \frac{1}{10} \]

### Step 4: Calculate the probabilities
Now we can calculate the probabilities for each outcome:
- \( P(1) = \frac{1}{10} \)
- \( P(2) = 3 \times \frac{1}{10} = \frac{3}{10} \)
- \( P(3) = \frac{1}{10} \)
- \( P(4) = 3 \times \frac{1}{10} = \frac{3}{10} \)
- \( P(5) = \frac{1}{10} \)
- \( P(6) = \frac{1}{10} \)

### Step 5: Identify the combinations for a total of 7
The combinations of the two dice that can yield a total of 7 are:
1. (1, 6)
2. (2, 5)
3. (3, 4)
4. (4, 3)
5. (5, 2)
6. (6, 1)

### Step 6: Calculate the probability of each combination
Now we calculate the probability for each of these combinations:
- For (1, 6): \( P(1) \times P(6) = \frac{1}{10} \times \frac{1}{10} = \frac{1}{100} \)
- For (2, 5): \( P(2) \times P(5) = \frac{3}{10} \times \frac{1}{10} = \frac{3}{100} \)
- For (3, 4): \( P(3) \times P(4) = \frac{1}{10} \times \frac{3}{10} = \frac{3}{100} \)
- For (4, 3): \( P(4) \times P(3) = \frac{3}{10} \times \frac{1}{10} = \frac{3}{100} \)
- For (5, 2): \( P(5) \times P(2) = \frac{1}{10} \times \frac{3}{10} = \frac{3}{100} \)
- For (6, 1): \( P(6) \times P(1) = \frac{1}{10} \times \frac{1}{10} = \frac{1}{100} \)

### Step 7: Sum the probabilities for total 7
Now we sum these probabilities:
\[ P(7) = \frac{1}{100} + \frac{3}{100} + \frac{3}{100} + \frac{3}{100} + \frac{3}{100} + \frac{1}{100} = \frac{14}{100} = \frac{7}{50} \]

### Step 8: Calculate \( p \) and \( \frac{1}{p} \)
We have \( p = \frac{7}{50} \). Therefore, \( \frac{1}{p} \) is:
\[ \frac{1}{p} = \frac{50}{7} \]

### Step 9: Find the greatest integer less than or equal to \( \frac{1}{p} \)
Calculating \( \frac{50}{7} \) gives approximately \( 7.14 \). The greatest integer less than or equal to \( 7.14 \) is:
\[ \lfloor \frac{50}{7} \rfloor = 7 \]

### Final Answer
Thus, the value of \( \lfloor \frac{1}{p} \rfloor \) is \( 7 \).

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