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Suppose there are two groups of objects. Group $1$ has $50$ unique objects, and Group $2$ has $100$ unique objects. These groups share $40$ identical objects (i.e. $40/50$ for group $1$ and $40/100$ for group $2$). Given that $20$ objects are selected from group $1$ and $15$ objects are selected from group $2$, what is the probability that $10$ objects will be identical between the two selections.

My main issue with this problem is that the two groups are different sizes with partial overlap. I'm not sure how to take that into account when I calculate some sort of cumulative probability for each selection. Further, even if a shared object is selected from each group, how do I account for the fact that the same shared object may not be selected from both groups? What I've been doing now is just simulating random selections, but I'd like to know if there is some probability calculation that's applicable.

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I think this can be solved by elementary means, if I understand the problem correctly. This feels to simple to be right, so do correct me if I'm wrong.

There are ${50 \choose 20}$ ways to choose the objects from group $1$, and ${100 \choose 15}$ ways to choose objects from group $2$. Hence, ${50 \choose 20} \cdot {100 \choose 15}$ ways to make the two choices.

How many ways are there to choose the objects with $10$ element overlap? First, we select the $10$ common elements in ${40 \choose 10}$ ways. Next, we choose the remaining $10$ out of the group $1$, only we need to keep track of how many are in the common part of the two groups. Say we want $k$ elements from there (and $10-k$ from outside). Then, there are $10 \choose 10 -k$ ways to select the ones from outside the common part, and $30 \choose k$ to choose the ones from within the common part, giving the total of ${10 \choose 10 -k} \cdot {30 \choose k}$. Finally, we want to choose the $5$ remaining elements in group $2$. There were $100$ elements there initially, but we already chose $10$ out of them, and there are $k$ that we don't want to choose (else, the overlap would be larger than $10$). This gives ${90 - k \choose 5}$ ways. It remains to muliply things out and sum over $k$, to get the total number of ways: $$ {40 \choose 10} \sum_{k=0}^{10} {10 \choose 10 -k} \cdot {30 \choose k} \cdot {90 - k \choose 5}$$ I am not sure if this can be simplified.

To get the probablity, divide this value by the number of ways to make the choices. The numerical value is $0.00690417...$.

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Thanks, I think it makes sense now. –  user77978 May 17 '13 at 16:29

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