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

Explanation

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

### Step 1: Define the Random Variable
Let \( X \) denote the number of doublets obtained in three throws of a pair of dice. The possible doublets are:
- (1, 1)
- (2, 2)
- (3, 3)
- (4, 4)
- (5, 5)
- (6, 6)

Since we are throwing the dice three times, \( X \) can take values 0, 1, 2, or 3.

### Step 2: Calculate the Probability of Getting a Doublet
The probability of getting a doublet (same number on both dice) in a single throw is:
\[ P(\text{doublet}) = \frac{6}{36} = \frac{1}{6} \]
The probability of not getting a doublet is:
\[ P(\text{not doublet}) = 1 - P(\text{doublet}) = 1 - \frac{1}{6} = \frac{5}{6} \]

### Step 3: Calculate the Probability for Each Value of \( X \)

#### Case 1: \( X = 0 \) (No doublets)
The probability of getting no doublets in three throws is:
\[ P(X = 0) = \left(\frac{5}{6}\right)^3 = \frac{125}{216} \]

#### Case 2: \( X = 1 \) (One doublet)
The probability of getting exactly one doublet can occur in three different ways (doublet on the first throw, second throw, or third throw):
\[ P(X = 1) = 3 \cdot \left(\frac{1}{6}\right) \cdot \left(\frac{5}{6}\right)^2 = 3 \cdot \frac{1}{6} \cdot \frac{25}{36} = \frac{75}{216} \]

#### Case 3: \( X = 2 \) (Two doublets)
The probability of getting exactly two doublets can occur in three different ways (doublets on any two of the three throws):
\[ P(X = 2) = 3 \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right) = 3 \cdot \frac{1}{36} \cdot \frac{5}{6} = \frac{15}{216} \]

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

### Step 4: Compile the Probability Distribution
Now we can summarize the probability distribution of \( X \):

\[ \begin{align*} X & : 0 \quad 1 \quad 2 \quad 3 \\ P(X) & : \frac{125}{216} \quad \frac{75}{216} \quad \frac{15}{216} \quad \frac{1}{216} \end{align*} \]

### Final Answer
The probability distribution of the number of doublets in three throws of a pair of dice is:
- \( P(X = 0) = \frac{125}{216} \)
- \( P(X = 1) = \frac{75}{216} \)
- \( P(X = 2) = \frac{15}{216} \)
- \( P(X = 3) = \frac{1}{216} \)