Question

A secretary introduces 9 letters in 9 envelopes. Let X be the random variable representing the number of letters that entered the correct envelope. How you find `E(X)` and `var(X)`?

Answer 1

`E(X) = var(X) = 1`

Explanation:

`P(X=k) = (1/(k!)) sum_{i=0}^{9-k} (-1)^i/(i!)`

`=> P(X=0) = 1-1+1/2-1/6+1/24-1/120+1/720-1/5040+1/40320-1/362880 = 0.367879`
`=> P(X=1) = P(X=0) + 1/362880 = 0.367882`
`=> P(X=2) = (1/2-1/6+1/24-1/120+1/720-1/5040)/2 = 0.183929`
`=> P(X=3) = (1/2-1/6+1/24-1/120+1/720)/6 = 0.061343`
`=> P(X=4) = (1/2-1/6+1/24-1/120)/24 = 0.015278`
`=> P(X=5) = (1/2-1/6+1/24)/120 = 0.003125`
`=> P(X=6) = (1/2-1/6)/720 = 0.000463`
`=> P(X=7) = (1/2)/5040 = 0.000099`
`=> P(X=8) = 0`
`=> P(X=9) = 1/(9!) = 0.000003`

`=> E(X) = sum_{j=0}^{j=9} j (1/(j!)) sum_{i=0}^{9-j} (-1)^i/(i!)`

`= sum_{j=1}^{j=9} (1/((j-1)!)) sum_{i=0}^{9-j} (-1)^i/(i!)`

`= 0.367882 + 2*0.183929 + 3*0.061343 + 4*0.015278 +`
`5*0.003125 + 6*0.000463 + 7*0.000099 + 9*0.000003`
`= 1.000004`
`= 1`

`=> var(X) = E(X^2) - (E(X))^2`

`= 0.367882 + 4*0.183929 + 9*0.061343 + 16*0.015278 +`
`25*0.003125 + 36*0.000463 + 49*0.000099 + 81*0.000003`
`- 1.000004^2`
`= 1`