2020 AMC 10B Problems/Problem 11
Problem
Ms. Carr asks her students to read any 5 of the 10 books on a reading list. Harold randomly selects 5 books from this list, and Betty does the same. What is the probability that there are exactly 2 books that they both select?
Solution 1
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.
~quacker88
Solution 2
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 .
Solution 3
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, .
~BakedPotato66
Video Solution
~IceMatrix
~savannahsolver
https://youtu.be/wopflrvUN2c?t=118
~ pi_is_3.14
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AMC 10 Problems and Solutions |
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