I am having trouble calculating the following question: There are 42 people in different groups. Group A have 16 people. Group B have 11 people. Group C have 21 people. There are 6 people that are both in Group A and Group B. Suppose 5 people are randomly selected from 3 groups. calculate the probability that at least 1 people is from group A and at least one from B.
I am having trouble deal with the overlapping elements while calculating the combinations.
S = 42, A = 16, B = 11, A∩B = 6, C = 21
My first attempt was this: (16C1 x 11C1 x 15C3)/42C5 Then I realize that the overlapping elements could be an issue. I tried to separate it into three parts. That is, calculate the combinations without dealing with the overlapping elements. That is: ((16C1 x 5C1 x 21C3)+(10C1 x 11C1 x 21C3)+(10C0 x 5C0 x 6C2 x 21C3))/42C5 = 0.32
Then I realized that I forgot the "at least" condition So I tried this:
let A = 10, B = 5, C = 6, and D = 21
the probability of at least 1A1B, 1A1C, 1B1C, 2C
Pr = 1 - P[0A0B0C5D] - P[1A0B0C4D] - P[2A0B0C3D] - P[3A0B0C2D] - P[4A0B0C1D] - P[5A0B0C0D] - P[0A1B0C4D] - P[0A2B0C3D]- P[0A3B0C2D] - P[0A4B0C1D] - P[0A5B0C0D] - P[0A0B1C4D] = 70.465
not sure if it is correct.
Taking that one with label A and one with label B, which may be a single card labelled both A and B is mandatory,
groups that aren't "good" are those drawn from $16 A's$ and irrelevant $15$ + those drawn from $11 B's$ and irrelevant $15$
Thus $Pr = 1 - \dfrac{\binom{31}5 +\binom{26}5}{\binom{42}5}$
ADDED
Re your further comments, "bad" combos will be
$A\; B\; AB\; C$
$0\;\; 0\;\;\; 0\;\;\; 5$
$0\;\; 0\;\;\; 1\;\;\; 4$
$0\;\; 1\;\;\; 0\;\;\; 4$
$1\;\; 0\;\;\; 0\;\;\; 4$