Working out the probability of at least 4 out 7 Currently I have 15 things in one group and 3330 in another. Both groups are put together and I pick 7 things of them at random, what would the odds be that at least 4 out of 7 would be from the group with 3330 members
i know how to work out the the probability of getting 4 out of 7 from the 3330 group is given by
(3330/4)(15/3)(3345/7)=2.519×10−6
the issue i have is while this is ok, i want  to know the probability of picking at least 4 from the 3330 group and to include 4,5,6 and 7 possibilities and not 4 from 1 group and 3 from another
NB: the probability of getting all 7 out of 7 from the 3330 group is
(3330/7)(15/0)(3345/7)=0.9690
how can i do this ?
 A: You just have to notice that if $X$ is the number of the 7 chosen items which come from the larger group, then
$$
P(X\geq 4)=P(X=4)+P(X=5)+P(X=6)+P(X=7),
$$
since these are disjoint events.  You've already computed $P(X=4)$ and $P(X=7)$; now, you just need to do the same for $X=5$ and $X=6$.
I should add, though, that there is an issue with your computed $P(X=4)$. The event $X=4$ means that you are choosing any four elements from the large group, and any three elements from the small group; so,
$$
P(X=4)=\frac{\binom{3330}{4}\cdot\binom{15}{3}}{\binom{3345}{7}}.
$$
(This might be what you intended, but I can't tell from your notation.)
A: No. of ways to pick $j$ things from the $3330$ group (let me refer this to as $B$) = $\binom{3330}{j}$. 
No. of ways to pick $7-j$ from the $15$ group (let me refer this to as $A$) = $\binom{15}{7-j}$. 
Then the probability of picking $j$ from $B$ and $7-j$ from $A$ is $\dfrac{\binom{3330}{j}\binom{15}{7-j}}{\binom{3345}{7}}$.
Now to pick at least $4$ from group $B$. You need to add the probabilities.
$$\sum_{j=4}^{j=7}\frac{\binom{3330}{j}\binom{15}{7-j}}{\binom{3345}{7}}$$
