# How to get probability of getting only those items from box which you did not pick before?

I have a box which contains $$n$$ unused items (all items in the box are unused). From them I randomly pick $$k$$, where $$k < n$$ items. Those $$k$$ items became used when I picked them, and then I put them back into the box. How to find the probability that second time all $$k$$ items I pick are going to be unused?(not those which I already picked before)

I already know that all possible ways of picking $$k$$ from $$n$$ is: $$\frac{n!}{k!(n-k)!}$$ and that possible ways of picking $$k$$ from $$(n-k)$$ is : $$\frac{(n-k)!}{k!((n-k)-k)!}$$

• Welcome to Math.SE. Take a look at How to ask a good question at Math.SE. To avoid downvotes and closing you should add your own efforts to the question by means of an edit (not a comment), and tell us where you got stuck.Hint: How many ways are there to pick $k$ out of $n$ items? How many ways are there to pick $k$ items out of $n-k$ (unused) items? Commented Oct 26, 2019 at 8:24

## 1 Answer

There are indeed $$\binom{n}{k}$$ ways of picking $$k$$ items out of $$n$$.

The number of possible ways of picking $$k$$ from $$n-k$$ is $$\binom{n-k}{k}$$ (so not $$\frac{n!}{k!(n-2k)!}$$).

These "ways" are equiprobable so the event of picking the second time $$k$$ items that are not picked the first times has probability:$$\frac{\binom{n-k}{k}}{\binom{n}{k}}$$

• The numerator is the number of ways that $k$ cards can be picked from the $n-k$ cards that where not picked the first time: $\binom{n-k}{k}=\frac{(n-k)!}{k!(n-2k)!}$. You write wrongly that there are $\frac{n!}{k!(n-2k)!}$ possibilities. E.g. let $n=4$ and $k=2$. Then there is only $1=\binom{4-2}{2}$ way to pick the $2$ cards that are not picked the first time. Not $\frac{4!}{2!0!}$ ways. Commented Oct 27, 2019 at 15:53