Find the mean of the probability distribution of the number of doublets in three throws of a pair of dice.

Answer

To find the mean of the probability distribution of the number of doublets in three throws of a pair of dice, we will follow these steps:

### Step 1: Identify the possible outcomes
When throwing a pair of dice, the possible doublets (same number on both dice) are:
- (1,1)
- (2,2)
- (3,3)
- (4,4)
- (5,5)
- (6,6)

Thus, there are 6 doublets out of a total of 36 possible outcomes when throwing two dice.

### Step 2: Calculate the probabilities
The probability of getting a doublet (P) is:
\[ P = \frac{6}{36} = \frac{1}{6} \]

The probability of not getting a doublet (Q) is:
\[ Q = 1 - P = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 3: Determine the number of doublets in three throws
We can have 0, 1, 2, or 3 doublets in three throws. We will calculate the probability for each case.

1. **Case 1: 0 doublets**
\[ P(X=0) = Q^3 = \left(\frac{5}{6}\right)^3 = \frac{125}{216} \]

2. **Case 2: 1 doublet**
The probability of getting exactly 1 doublet can occur in three different ways (doublet in one throw, no doublet in the other two):
\[ P(X=1) = \binom{3}{1} \cdot P \cdot Q^2 = 3 \cdot \frac{1}{6} \cdot \left(\frac{5}{6}\right)^2 = 3 \cdot \frac{1}{6} \cdot \frac{25}{36} = \frac{75}{216} \]

3. **Case 3: 2 doublets**
The probability of getting exactly 2 doublets:
\[ P(X=2) = \binom{3}{2} \cdot P^2 \cdot Q = 3 \cdot \left(\frac{1}{6}\right)^2 \cdot \frac{5}{6} = 3 \cdot \frac{1}{36} \cdot \frac{5}{6} = \frac{15}{216} \]

4. **Case 4: 3 doublets**
The probability of getting 3 doublets:
\[ P(X=3) = P^3 = \left(\frac{1}{6}\right)^3 = \frac{1}{216} \]

### Step 4: Compile the probability distribution
We can summarize the probabilities as follows:
- \( P(X=0) = \frac{125}{216} \)
- \( P(X=1) = \frac{75}{216} \)
- \( P(X=2) = \frac{15}{216} \)
- \( P(X=3) = \frac{1}{216} \)

### Step 5: Calculate the mean
The mean (expected value) of the probability distribution is calculated using the formula:
\[ E(X) = \sum (x_i \cdot P(X=x_i)) \]

Calculating the mean:
\[ E(X) = 0 \cdot \frac{125}{216} + 1 \cdot \frac{75}{216} + 2 \cdot \frac{15}{216} + 3 \cdot \frac{1}{216} \]
\[ = 0 + \frac{75}{216} + \frac{30}{216} + \frac{3}{216} \]
\[ = \frac{75 + 30 + 3}{216} = \frac{108}{216} = \frac{1}{2} \]

### Conclusion
The mean of the probability distribution of the number of doublets in three throws of a pair of dice is:
\[ \boxed{\frac{1}{2}} \]