Difference between revisions of "2020 AMC 10B Problems/Problem 11"
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==Problem== | ==Problem== | ||
− | Ms. Carr asks her students to | + | Ms. Carr asks her students to read any <math>5</math> of the <math>10</math> books on a reading list. Harold randomly selects <math>5</math> books from this list, and Betty does the same. What is the probability that there are exactly <math>2</math> books that they both select? |
<math>\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}</math> | <math>\textbf{(A)}\ \frac{1}{8} \qquad\textbf{(B)}\ \frac{5}{36} \qquad\textbf{(C)}\ \frac{14}{45} \qquad\textbf{(D)}\ \frac{25}{63} \qquad\textbf{(E)}\ \frac{1}{2}</math> | ||
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~BakedPotato66 | ~BakedPotato66 | ||
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+ | ==Solution 4== | ||
+ | Say that there is a set <math>S</math> <math>{a,b,c,...,j}</math> which represent the 10 books. We know that intersection of the subsets that Bob and Alice choose has 2 elements. Thus there are <math>\dbinom{10}{2}</math> ways for that to happen. Out of the 8 remaining in the set <math>S</math>, Bob and Alice both need to choose 3 different elements = <math>\dbinom{8}{3} * \dbinom{5}{3}</math>. Thus the number of ways this works is <math>\dbinom{10}{2} * \dbinom{8}{3} * \dbinom{5}{3}</math>. The denominator for the fraction is just <math>\dbinom{10}{5} * \dbinom{10}{5}</math>. Simplifying, you get <math>\boxed{\frac{25}{63}}</math>. | ||
+ | |||
+ | ~YBSuburbanTea | ||
+ | |||
+ | ==Video Solution (HOW TO CRITICALLY THINK!!!)== | ||
+ | https://youtu.be/wwKJBG5zCAE | ||
+ | |||
+ | ~Education, the Study of Everything | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | https://youtu.be/wopflrvUN2c?t=118 | ||
+ | -Sohil Rathi | ||
+ | ~Juicer100 | ||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/t6yjfKXpwDs | https://youtu.be/t6yjfKXpwDs | ||
− | + | ||
https://youtu.be/RbKdVmZRxkk | https://youtu.be/RbKdVmZRxkk | ||
~savannahsolver | ~savannahsolver | ||
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==See Also== | ==See Also== | ||
{{AMC10 box|year=2020|ab=B|num-b=10|num-a=12}} | {{AMC10 box|year=2020|ab=B|num-b=10|num-a=12}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:32, 31 August 2024
Contents
Problem
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?
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
Solution 4
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
.
~YBSuburbanTea
Video Solution (HOW TO CRITICALLY THINK!!!)
~Education, the Study of Everything
Video Solution
https://youtu.be/wopflrvUN2c?t=118 -Sohil Rathi ~Juicer100
Video Solution
~savannahsolver
See Also
2020 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.