Given $n$ blue letters and $n$ pink letters (in total $2n$ letters) to place in $n$ envelopes. on each envelop there's an address written, and the letters also have $n$ different addresses. In each envelope we randomly choose 2 letters - 1 pink and 1 blue.
Let $X$ be the number of envelopes which has at least one correct letter in them. what is the Expected value of $X$?
I understand that if we have $n$ letters, then $E[X] = n\cdot \frac{1}{n} =1$
I want to find the probability of putting 2 correct letters in an envelope, and to use the Complement of it to find the probability of at least 1 correct letter, but the probability for the $j$-envelope is depended on the previous $j-1$ envelopes and this is where I get lost.
We can use linearity of expectation here. Let $X_i$ be $1$ or $0$ according to whether the $i$-th envelope has at least one correct letter in it or not.
Then $E[X_i]=P($ at least one correct letter in envelope $i)=\frac1n+\frac1n-\frac{1}{n^2}$ (Using inclusion-exclusion on pink letter correct or blue letter correct.)
We're interested in $E[X_1+\cdots+X_n]$. But by linearity of expectation this is equal to
$E[X_1]+\cdots+E[X_n]=n\cdot \left( \frac1n+\frac1n-\frac{1}{n^2}\right)=2-\frac1n$