Sum of Poisson Distribution, why is my solution incorrect

Question

From SOA #212:

The number of days an employee is sick each month is modeled by a Poisson distribution with mean $1$. The numbers of sick days in different months are mutually independent. Calculate the probability that an employee is sick more than two days in a three-month period.

I calculated the probabilities as follows

$${3\choose1}P(X=0)P(X=0)P(X\ge2) \ + $$ $${3\choose2}P(X=0)P(X\ge1)P(X\ge1) \ + $$ $${3\choose3}P(X\ge1)P(X\ge1)P(X\ge1) $$

$$=P(\text{employee is sick more than two days in a three-month period})$$

This should come out to $${3\choose1}\left({1^0e^{-1}\over0!}\right)\left({1^0e^{-1}\over0!}\right)\left(1- {1^0e^{-1}\over0!}-{1^1e^{-1}\over1!}\right) \ + $$

$${3\choose2}\left({1^0e^{-1}\over0!}\right)\left(1- {1^0e^{-1}\over0!}\right)\left(1- {1^0e^{-1}\over0!}\right) \ + $$

$${3\choose3}\left(1- {1^0e^{-1}\over0!}\right)\left(1- {1^0e^{-1}\over0!}\right)\left(1- {1^0e^{-1}\over0!}\right) \ $$

$$=.8005$$

This is incorrect, and I know how they calculated the correct answer, $.577$, using the sum of the independent Poisson variables, but I would like to know what is wrong with what I did.

Answer 1

You've computed the probability that the employee is sick on two or more days. The problem asks for the probability that the employee is sick on more than two days (that is, at least three days).

Answer 2

Your calculation does not match mine. It is structurally different. We want

$1-P(X\leq 2)=1-P(X=2)-P(X=1)-P(X=0)$

For $X=2$ we have several combinations

$110 \quad 200$

$011\quad 020$

$101\quad 002$

$X=1:$ $100,$ $\ 010,$$ \ 001$

$X=0$: $000$

Therefore the calculation is $1-7\cdot (1/e)^3-3\cdot (1/(2\cdot e))\cdot (1/e)^2=0.57681\approx 0.577$

Answer 3

We don't need to consider combinations because $X$ is also Poisson-distributed with mean $3$, making the answer $1-(1+3+3^2/2)/e^3=1-17/(2e^3)\approx 0.577$.