Difference between revisions of "2020 AMC 10B Problems/Problem 11"
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==Problem== | ==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? | + | 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> | ||
| − | ==Solution== | + | ==Solution 1== |
We don't care about which books Harold selects. We just care that Betty picks <math>2</math> books from Harold's list and <math>3</math> that aren't on Harold's list. | We don't care about which books Harold selects. We just care that Betty picks <math>2</math> books from Harold's list and <math>3</math> that aren't on Harold's list. | ||
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==Solution 2== | ==Solution 2== | ||
| − | We can analyze this as two containers with <math>10</math> balls each, with the two people grabbing <math>5</math> balls each. First, we need to find the probability of two of the balls being the same among five: <math>\frac{1}{3}\frac{4}{9}\frac{3}{8}\frac{5}{7}\frac{2}{4}</math>. After that we must | + | We can analyze this as two containers with <math>10</math> balls each, with the two people grabbing <math>5</math> balls each. First, we need to find the probability of two of the balls being the same among five: <math>\frac{1}{3} \cdot \frac{4}{9} \cdot \frac{3}{8} \cdot \frac{5}{7} \cdot \frac{2}{4}</math>. After that we must multiply this probability by <math>{5 \choose 2}</math>, for choosing the 2 balls that are the same chosen among |
| + | 5 balls. The answer will be <math>\frac{5}{126}*10 = \frac{25}{63}</math>. <math>\boxed{\textbf{(D) }\frac{25}{63}}</math> | ||
==Solution 3== | ==Solution 3== | ||
| − | + | Firstly, we know that the denominator will be <math>\dbinom{10}{5} \cdot \dbinom{10}{5}</math>. 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 <math>\dfrac{\binom{10}{2} \cdot \binom{8}{3} \cdot \binom{5}{3}}{\tbinom{10}{5} \cdot \tbinom{10}{5}} = \dfrac{45 \cdot 56 \cdot 10}{252 \cdot 252}</math>. 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, <math>\tbinom{10}{2}</math> is <math>5 \cdot 3 \cdot 3</math>). After doing so, we get, <math>\boxed{\textbf{(D) } \dfrac{25}{63}}</math>. | |
| + | |||
| + | ~BakedPotato66 | ||
| + | |||
| + | ==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>. | ||
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| + | ~YBSuburbanTea | ||
| + | |||
| + | ==Solution 5 (similar to 3 but way faster & cheese) == | ||
| + | Following the same method as solution 3, you end up getting <math>\dfrac{\binom{10}{2} \cdot \binom{8}{3} \cdot \binom{5}{3}}{\tbinom{10}{5} \cdot \tbinom{10}{5}} = \dfrac{45 \cdot 56 \cdot 10}{252 \cdot 252}</math>. However you notice that there are two <math>5</math>s in the numerator which can't cancel out. The only answer with two <math>5</math> in the numerator is <math>\boxed{\textbf{(D) }\frac{25}{63}}</math>. | ||
| + | |||
| + | ~ neeyakkid23 | ||
| + | |||
| + | ==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 | ||
Latest revision as of 12:12, 4 November 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
Solution 5 (similar to 3 but way faster & cheese)
Following the same method as solution 3, you end up getting . However you notice that there are two s in the numerator which can't cancel out. The only answer with two in the numerator is .
~ neeyakkid23
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. 
