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Kind of embarrassed to admit that I am at a loss with this problem, but here it goes...

I have ten items in a set. Three items are selected from the set, then the set is put back to normal; it is reset to ten items in the set. Three items are again selected from the full set. What is the probability that, on the second selection, an item was selected that was also selected in the first selection?

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    I assume you mean *at least one item* was selected in both sets. It is far easier to work with the opposite problem. What is the probability that none of the items selected match? To do this, think of it this way. All of the items start off colored black. The three you select the first time, paint them red and put them back in the set. So, you have three red items and seven black items. What is the probability that you pick only black items the next time around? What does this imply about your original problem? – 2017-02-21
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    Re: "What is the probability that, on the second selection, **an item** was selected that was also selected in the first selection?" It is ambiguous if "an item" means "exactly one" or "at least one." The two interpretations have different answers. – 2017-02-21
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    To clarify; in the first round of picks three items are chosen from a set, then in the second round three items are picked from the full set again. I want to known the mathematics behind the probability that **at least one item** picked in the first round (3/10) is the same as an item picked in the second round (3/10). – 2017-02-22

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The easiest way to answer any problem like this is to find the probability of the complement. You're essentially asking "what is the probability that at least one item...". At least one should always ring alarm bells, and you should instead find the probably that none of the items do what you want.

With that in mind, $\mathbb{P}[\text{No common items}]=\frac{7}{10}\frac{6}{9}\frac{5}{8}=\frac{7}{24}$.

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    I'm still confused unfortunately. :/ – 2017-02-22