To solve the problem of finding the mean and variance of the number of doublets when rolling a pair of fair dice twice, we will follow these steps:
### Step 1: Understand the Concept of Doublets
Doublets occur when both dice show the same number. For example, (1,1), (2,2), (3,3), (4,4), (5,5), and (6,6) are all doublets. There are a total of 6 doublets possible.
### Step 2: Calculate the Probability of Getting a Doublet
When rolling two dice, the total number of outcomes is \(6 \times 6 = 36\). The probability of getting a doublet (D) is given by:
\[
P(D) = \frac{\text{Number of doublets}}{\text{Total outcomes}} = \frac{6}{36} = \frac{1}{6}
\]
The probability of not getting a doublet (D') is:
\[
P(D') = 1 - P(D) = 1 - \frac{1}{6} = \frac{5}{6}
\]
### Step 3: Define the Random Variable
Let \(X\) be the random variable representing the number of doublets in two rolls of the dice. The possible values of \(X\) are 0, 1, or 2.
### Step 4: Calculate the Probabilities for Each Value of \(X\)
1. **Probability that \(X = 0\)** (no doublets):
\[
P(X = 0) = P(D') \times P(D') = \left(\frac{5}{6}\right) \times \left(\frac{5}{6}\right) = \frac{25}{36}
\]
2. **Probability that \(X = 1\)** (one doublet):
There are two scenarios: doublet on the first roll and not on the second, or not on the first and doublet on the second.
\[
P(X = 1) = P(D) \times P(D') + P(D') \times P(D) = \left(\frac{1}{6} \times \frac{5}{6}\right) + \left(\frac{5}{6} \times \frac{1}{6}\right) = \frac{5}{36} + \frac{5}{36} = \frac{10}{36}
\]
3. **Probability that \(X = 2\)** (two doublets):
\[
P(X = 2) = P(D) \times P(D) = \left(\frac{1}{6}\right) \times \left(\frac{1}{6}\right) = \frac{1}{36}
\]
### Step 5: Create the Probability Distribution Table
\[
\begin{array}{|c|c|}
\hline
X & P(X) \\
\hline
0 & \frac{25}{36} \\
1 & \frac{10}{36} \\
2 & \frac{1}{36} \\
\hline
\end{array}
\]
### Step 6: Calculate the Mean of \(X\)
The mean (expected value) \(E(X)\) is calculated as:
\[
E(X) = \sum (X \cdot P(X)) = 0 \cdot \frac{25}{36} + 1 \cdot \frac{10}{36} + 2 \cdot \frac{1}{36}
\]
\[
E(X) = 0 + \frac{10}{36} + \frac{2}{36} = \frac{12}{36} = \frac{1}{3}
\]
### Step 7: Calculate the Variance of \(X\)
To find the variance, we first need \(E(X^2)\):
\[
E(X^2) = \sum (X^2 \cdot P(X)) = 0^2 \cdot \frac{25}{36} + 1^2 \cdot \frac{10}{36} + 2^2 \cdot \frac{1}{36}
\]
\[
E(X^2) = 0 + \frac{10}{36} + \frac{4}{36} = \frac{14}{36}
\]
Now, the variance \(Var(X)\) is given by:
\[
Var(X) = E(X^2) - (E(X))^2 = \frac{14}{36} - \left(\frac{1}{3}\right)^2
\]
\[
Var(X) = \frac{14}{36} - \frac{1}{9} = \frac{14}{36} - \frac{4}{36} = \frac{10}{36}
\]
### Final Answers
- Mean of the number of doublets: \(\frac{1}{3}\)
- Variance of the number of doublets: \(\frac{10}{36}\)
