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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?

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up vote 0 down vote accepted

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:

$$ \frac{Y}{X} = \frac{18^6-17^6}{18^6} = 1 -\left( \frac{17}{18}\right)^6$$

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I don't think that's a very good direction. It could require much work. I always fond it useful to think of a simplified version:

Suppose we pick 1 loaf. What are the chances of it being blueberry?

Suppose we pick 2 loaves. What are the chances of not having a blueberry?

3 loaves?

4 loaves?

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6 loaves of bread are selected with repetitions allowed. Therefore, there are $18^6$ possible lists of bread without any restraints. Now if we included the condition that at least one of them is blueberry, it would be easier to approach the question from the converse. The opposite of at least one of them being blueberry is none being blueberry. There are $17^6$ possible choices in which none of the six bread is blueberry, and therefore $18^6-17^6$ choices where at least one is blueberry. So now we have the answer,

$$(18^6-17^6)/18^6 = 0.29$$

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