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?

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 $\frac{219}{784}=0.2793$
