Probability question with combinations of different types of an item Suppose a bakery has 18 varieties of bread, one of which is blueberry bread. If a half dozen loafs of bread are selected at random (with repetitions allowed), then what is the probability that at least one of the loafs of blueberry bread will be included in the selection.
I started off by determining that if we picked up at least one blueberry bread from the start, then we could find the total ways by finding:
$C_{R}(n, r)$
Plugging in for $n$ and $r$ I calculated that the number of ways would be:
$C_{R}(18, 5) = C(18+5-1,5) = 26,344$ ways.
Am I on the right track, and how would I go about finding the probability in this case?

You can have $18^6$ possible different "half dozen loafs of bread" since you are dealing with ordering $18$ element with repetition in group of 6. Let $X = 18^6$.
Now you have to consider the cases where there is at least one blueberry. Let the number of these ordering with repetition be $Y$.
Let $Z$ be the number of ordering with repetition in which there is NO blueberry.
It's clear that $X = Y + Z$ and then $Y = X - Z$.
Now we work on $Z$. To obtain $Z$ we can think to remove the blueberry among all the possibility and then count all the possible half dozen that we can have.
Now, since we have $17$ different types of bread, then we have to consider the ordering of $17$ element with repetition in group of 6. So $Z = 17^6$. And
$$Y = 18^6 - 17^6$$
$Y$ are all the half dozen with at least $1$ blueberry.
Finally, the probability is $Y/X$ (the number of interesting cases over the number of all the possible cases). We obtain:
$$ \frac{Y}{X} = \frac{18^6-17^6}{18^6} = 1 -\left( \frac{17}{18}\right)^6$$
