Answer will be 1.
There can be 2 ways to do it.
Method 1:
Let, $X_i$ be a indicator random variable which is defined as:
$X_i=1,$ if $i^{th}$ letter goes to its corresponding address
$X_i=0$, otherwise
Since, $3$ letters can be placed into 3 addressed envelopes in $3! = 6$ ways.
So, Mean = $E(X) = E(X_1 + X_2+ X_3) = E(X_1)+E(X_2)+ E(X_3)= 1* \frac{2}{6} + 1* \frac{2}{6} + 1* \frac{2}{6} = 1$
Now, Variance = $E(X^2) – E(X)^2 = E(X^2) – 1$
Now, $E(X^2) = E((X_1+X_2+X_3)^2) = E(X_1^2+X_2^2+ X_3^2+ 2X_1X_2+2X_1X_3+2X_2X_3)$
Using Linearity of Expectation,
$E(X^2) = E(X_1^2)+ E(X_2^2)+ E(X_3^2)+ 2E(X_1X_2)+2E(X_1X_3)+ 2E(X_2X_3)$
$X_1^2$ means first letter goes to its corresponding address (only one point is fixed)
$X_1X_2$ means first letter goes to its corresponding address as well as second letter goes to its corresponding address (two points are fixed)
So, $E(X^2) = 1^2* \frac{2}{6} + 1^2* \frac{2}{6} + 1^2* \frac{2}{6} + 2*1^2 *\frac{1}{6} + 2*1^2 *\frac{1}{6} + 2*1^2 *\frac{1}{6}$
So, $E(X^2) = 1+ 1=2$
Hence, Variance = $2-1 = 1$
Method 2: it is simple.
Let $X$ be a random variable which denotes number of letters goes into its corresponding addresses.
So, $P(X=1) = \frac{3}{6}$
$P(X=2) = 0$
$P(X=3) = \frac{1}{6}$
Now, Mean = $E(X) = \Sigma x_i P(X=x_i) = 1*\frac{3}{6}+2*0+3*\frac{1}{6} = 1 $
And Variance= $\Sigma x_i^2 P(X=x_i) – 1 = 1^2*\frac{3}{6} +0 + 3^2 *\frac{1}{6} -1 = 1$