Determine the probability that each of the 8 members serves on at least one of the three committees. A certain group has 8 members. In January, 3 members are selected at random to serve on a  committee. In February, 4 members are selected at random and independently of the first selection to serve on another committee.  In March, 5 members are selected at random and independently of the previous 2 selections to serve on a third committee. Determine the probability that each of the 8 members serves on at least one of the three committees.
So I am really unsure about this problem, I put the committees choice in the denominator, to represent the total choices. My reasoning for the numerator were you can choose any 3 people from the 8, then from the remaining 5 who haven't had positions you have to choose 4 and then there is 1 left who hasn't had a position so he has to be one of the 5 selected in March, and the remaining 4 can be any from the 
$\frac{{8 \choose 3}{5 \choose 4}{1 \choose 1}{7 \choose 4}}{{8 \choose 3}{8 \choose 4}{8 \choose 5}}$
So I need to use the inclusion exclusion principle?
 A: Hint 1: Simplify the problem.
Five members serve on the March committee and three do not.  What is the probability that there's at least one of those three who did not serve on either of the other two committees?  What is the complement of this?
Hint 2: Use the Principle of Inclusion and Exclusion.
A: WLOG, assume that the members $1$ to $5$ belong to the third committee.
The probability that $4$ of those members belong to the second committee, is
 $\frac{\binom{5}{4}}{\binom{8}{4}}=\frac{1}{14}$. In this case, all the remaining
 members must belong to the first committee, so $\frac{1}{14}\times\frac{1}{\binom{8}{3}}=\frac{1}{784}$ is the probability in this case.
The probability that $3$ of those members belong to the second committee, is
 $\frac{\binom{5}{3} \times \binom{3}{1}} {\binom{8}{4}}=\frac{3}{7}$. In
 this case, the $2$ remaining members must belong to the first committee, so
 $\frac{3}{7}\times \frac{6}{\binom{8}{3}}=\frac{9}{196}$ is the 
 probability in this case.
The probability that $2$ of those members belong to the second committee, is
 $\frac{\binom{5}{2} \times \binom{3}{2}} {\binom{8}{4}}=\frac{3}{7}$. In this
 case, the remaining member must belong to the first committee, so $\frac{3}{7}\times\frac{3}{8}=\frac{9}{56}$ is the probability in this case.
Finally, the probability that $1$ of those members belongs to the second committee
 is $\frac{1}{14}$. In this case, all members already belong to at least one 
 committee.
In total, the probability is $\frac{219}{784}=0.2793$
