What is the probability that at least one group is not represented? A group of 20 persons from Region A is constituted from 7 persons from group X, 10 persons from group Y and 3 persons from group Z. 5 persons are randomly selected to form a group B. What is the probability that there is at least one group not represented by group B? So far what i have is this: $$ \frac{\binom{17}{5}+\binom{10}{5}+\binom{13}{5}}{\binom{20}{5}}= \frac{7727}{15504}= 0.49839 $$ is this the right answer?

Your term $\binom{17}{5}$ counts all choices that leave out Z. The term $\binom{10}{5}$ counts all choices that leave out Y. Their sum double-counts the choices that leave out Z and Y, that is, consist of members of X alone. So we must subtract $\binom{7}{5}$ to get rid of the double-counting.

Your sum also double-counts the choices where only Y is represented. So we must also subtract $\binom{10}{5}$ from the sum. There are no choices where only Z is represented, so no further adjustment is necessary.

**Remark:** The strategy was basic Inclusion/Exclusion, deliberate overcounting followed by adjustment.