If three letters are placed in three addressed envelopes then the mean and variance of X where X denotes the number of correct despatches.

Options

A. 1,1
B. 1,2
C. 2,1
D. 2,2

Answer

To solve the problem of finding the mean and variance of \( X \), where \( X \) denotes the number of correct dispatches when three letters are placed in three addressed envelopes, we can follow these steps:

### Step 1: Identify Possible Values of \( X \)
The number of correct dispatches \( X \) can take the values:
- \( 0 \): No letters are correctly dispatched.
- \( 1 \): One letter is correctly dispatched.
- \( 2 \): Two letters are correctly dispatched.
- \( 3 \): All three letters are correctly dispatched.

### Step 2: Calculate Total Ways to Dispatch Letters
The total number of ways to dispatch the three letters into three envelopes is given by the factorial of the number of letters, which is:
\[ 3! = 6 \]

### Step 3: Calculate Probabilities for Each Value of \( X \)
1. **Probability of \( X = 3 \)** (All letters correctly dispatched):
- There is only 1 way for this to happen (all letters in the correct envelopes).
- Probability:
\[ P(X = 3) = \frac{1}{6} \]

2. **Probability of \( X = 2 \)** (Two letters correctly dispatched):
- If two letters are correct, the third letter must also be correct (since there is only one envelope left).
- Probability:
\[ P(X = 2) = 0 \]

3. **Probability of \( X = 1 \)** (One letter correctly dispatched):
- We can choose any one of the three letters to be correct, and the other two must be incorrect.
- There are 3 ways to choose which letter is correct.
- Probability:
\[ P(X = 1) = \frac{3}{6} = \frac{1}{2} \]

4. **Probability of \( X = 0 \)** (No letters correctly dispatched):
- The only arrangement where no letters are correctly dispatched is when all letters are in the wrong envelopes.
- There are 2 ways to arrange this (derangements).
- Probability:
\[ P(X = 0) = \frac{2}{6} = \frac{1}{3} \]

### Step 4: Summary of Probabilities
- \( P(X = 0) = \frac{1}{3} \)
- \( P(X = 1) = \frac{1}{2} \)
- \( P(X = 2) = 0 \)
- \( P(X = 3) = \frac{1}{6} \)

### Step 5: Calculate the Mean of \( X \)
The mean \( E(X) \) is calculated as follows:
\[ E(X) = \sum (x \cdot P(X = x)) = 0 \cdot \frac{1}{3} + 1 \cdot \frac{1}{2} + 2 \cdot 0 + 3 \cdot \frac{1}{6} \]
Calculating this gives:
\[ E(X) = 0 + \frac{1}{2} + 0 + \frac{3}{6} = \frac{1}{2} + \frac{1}{2} = 1 \]

### Step 6: Calculate the Variance of \( X \)
The variance \( Var(X) \) is calculated using the formula:
\[ Var(X) = E(X^2) - (E(X))^2 \]
First, we calculate \( E(X^2) \):
\[ E(X^2) = 0^2 \cdot \frac{1}{3} + 1^2 \cdot \frac{1}{2} + 2^2 \cdot 0 + 3^2 \cdot \frac{1}{6} \]
Calculating this gives:
\[ E(X^2) = 0 + \frac{1}{2} + 0 + \frac{9}{6} = \frac{1}{2} + \frac{3}{2} = 2 \]
Now substituting back into the variance formula:
\[ Var(X) = 2 - (1)^2 = 2 - 1 = 1 \]

### Final Results
- Mean \( E(X) = 1 \)
- Variance \( Var(X) = 1 \)