2020 AMC 10B Problems/Problem 11
Ms. Carr asks her students to read any of the books on a reading list. Harold randomly selects books from this list, and Betty does the same. What is the probability that there are exactly books that they both select?
We don't care about which books Harold selects. We just care that Betty picks books from Harold's list and that aren't on Harold's list.
The total amount of combinations of books that Betty can select is .
There are ways for Betty to choose of the books that are on Harold's list.
From the remaining books that aren't on Harold's list, there are ways to choose of them.
We can analyze this as two containers with balls each, with the two people grabbing balls each. First, we need to find the probability of two of the balls being the same among five: . After that we must multiply this probability by , for choosing the 2 balls that are the same chosen among 5 balls. The answer will be .
Firstly, we know that the denominator will be . To calculate the numerator, or successful events, we first find the number of ways both Betty and Harold can choose the same 2 books. Then we find the number of ways for Betty to choose 3 different other books and the number of ways for Harold to choose 3 different other books. That is . From here, do not multiply, instead cancel common factors, then simplify. In fact, to make this expression even more manageable, leave the combinations in simplest factored form (for example, is ). After doing so, we get, .
Say that there is a set which represent the 10 books. We know that intersection of the subsets that Bob and Alice choose has 2 elements. Thus there are ways for that to happen. Out of the 8 remaining in the set , Bob and Alice both need to choose 3 different elements = . Thus the number of ways this works is . The denominator for the fraction is just . Simplifying, you get .
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