We have $m$ groups, each of the $m$ groups has $a_i$ members. Because the first group ($a_1)$ is special, we want to calculate the probability of choosing $n$ members from the groups, such that there will be at least one member from the first group. From each group we can choose at most $a_i$ members, and we can also choose $0$ members from that group. All the events are independent and random and the order doesn't matter.
What I think for the solution
Let $P$ be our probability, we can define $P = \frac{\text{combinations having elements from first group}}{\text{total number of combinations} }$. We also define sum $S = a_1 + a_2 + a_3 + \dots + a_m$. The total number of combinations is $\binom{S}{n}$. But I'm stuck into calculating the number of combinations having elements from first group. I was thinking that for each $1 \leq i \leq a_1$ We should calculate $\binom{S-a_1}{n-i}$ and sum those results. But I think that I'm something wrong. Because I don't have much experience with calculating probability please give me some hints how to correct myself and solve this on easier way.
Let $p$ be the probability that no person from the first group is selected. Then the desired probability is $1 - p$.