Variance of X in envelope question

Question

Three letters are placed into three addressed envelopes randomly. A random variable X denotes the number of letters placed into corresponding envelopes. The variance of X is __________.

Comments

2 is the answer ?

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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= $\Sigma x_i^2 P(X=x_i) – 1 = 1^2*\frac{3}{6} +0 + 3^2 *\frac{1}{6} -1 = 1$

Answer 1

The variance of X would be 2/9. This is because, since each letter is placed randomly into one of the three addressed envelopes, the probability of any letter going into a specific envelope is 1/3. As the three letters are independent of each other and the probability of any letter going into a specific envelope is 1/3, the variance of X is the sum of the variances of the individual Bernoulli random variables. Each Bernoulli random variable has variance p(1-p) = (1/3)(1-1/3) = 1/3 * 2/3 = 2/9 . Therefore, the variance of X is 2/9

Comments

how 2/9 =1/3 ?