Twenty persons are attending a meeting in a hall, 8 of whom left the hall and have returned at least once so far. Four persons have just left and are coming back. This coincides with the selection of a subcommittee of 4 persons who will be chosen by everyone. What is the probability that all subcommittee members will have never left the hall?
Answer: 0.0125
My Approach
Let $E_{x}$ be the event that $x$ persons out of $4$ who are outside are from out of the ones that have gone back and came back at least once
Then out of 8 people, $8-x$ remain back for choosing the subcommittee. And out of remaining 16, $8+x$ persons would have never gone out.
Let $N\equiv$ the event that the selection of 4 people does not contain from the 16 who stayed back contains people who never went out.
Since $E_{i}$s are mutually exclusive and $\displaystyle\bigcup_{i=0}^{4}E_{i}=S$, the sample space, we have,
$P(N)=\displaystyle\sum_{x=0}^{4}P(N|E_{x})P(E_{x})$
Since $P(E_{x})=\displaystyle\frac{\binom{8}{x}\binom{12}{4-x}}{\binom{20}{4}}$
and $P(N|E_{x})=\displaystyle\frac{\binom{8+x}{4}}{\binom{16}{4}}$, we have
$P(N)=\displaystyle\sum_{x=0}^{4}\displaystyle\frac{\binom{8+x}{4}\binom{8}{x}\binom{12}{4-x}}{\binom{16}{4}\binom{20}{4}}$.
But this is giving me the answer 33/323 (Using PARI/GP CAS).
Please review my approach and suggest any corrections.