Difference between revisions of "2021 Fall AMC 12B Problems/Problem 23"
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==Solution 1== | ==Solution 1== | ||
− | There are <math>29</math> possible pairs of consecutive integers, namely <math>\{1,2\}, \{ | + | There are <math>29</math> possible pairs of consecutive integers, namely <math>p_1=\{1,2\}, \cdots, p_{29}=\{29,30\}</math>. |
− | + | Let <math>X_i</math> be a random variable with value <math>X_i=1</math> if <math>p_i</math> is part of the 5-element subset, and 0 otherwise. | |
− | The | + | The number of pairs of consecutive integers in a 5-element selection is given by the sum <math>X_1+\cdots + X_{29}</math>. By linearity of expectation, the expected value is equal to the sum of the <math>\mathbb{E}[X_i]</math>: |
− | + | <cmath>\mathbb{E}[X_1+\cdots +X_{29}]=\mathbb{E}[X_1]+\cdots + \mathbb{E}[X_{29}]</cmath> | |
− | + | To compute <math>\mathbb{E}[X_i]</math>, note that the number of selections for which <math>X_i=1</math> is <math>\binom{28}{3}</math> out of <math>\binom{30}{5}</math> possible selections. Thus<cmath>\mathbb{E}[X_i]=\frac{\binom{28}{3}}{\binom{30}{5}}= \frac 1{29}\cdot \frac 23, \quad \textrm{which implies} \quad \mathbb{E}[X_1+\cdots +X_{29}]= \boxed{\textbf{(A)}\ \frac{2}{3}}.</cmath> | |
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~kingofpineapplz | ~kingofpineapplz | ||
+ | ==Solution 2== | ||
Alternatively, using linearity of expectation on the chosen subset: there are <math>{5 \choose 2}</math> pairs of integers. There are 29 pairs of consecutive integers out of <math>{30 \choose 2}</math> possible pairs, thus the probability that any given pair is consecutive is <math>\frac{29}{{30 \choose 2}}</math>. By linearity of expectation, there are <math>\frac{{5 \choose 2}29}{{30 \choose 2}} = \frac{10 \cdot 29}{\frac{30 \cdot 29}{2}} = \boxed{\textbf{(A)}\ \frac{2}{3}}</math> such pairs on average. | Alternatively, using linearity of expectation on the chosen subset: there are <math>{5 \choose 2}</math> pairs of integers. There are 29 pairs of consecutive integers out of <math>{30 \choose 2}</math> possible pairs, thus the probability that any given pair is consecutive is <math>\frac{29}{{30 \choose 2}}</math>. By linearity of expectation, there are <math>\frac{{5 \choose 2}29}{{30 \choose 2}} = \frac{10 \cdot 29}{\frac{30 \cdot 29}{2}} = \boxed{\textbf{(A)}\ \frac{2}{3}}</math> such pairs on average. | ||
Revision as of 12:01, 9 June 2022
Problem
What is the average number of pairs of consecutive integers in a randomly selected subset of distinct integers chosen from the set ? (For example the set has pairs of consecutive integers.)
Solution 1
There are possible pairs of consecutive integers, namely . Let be a random variable with value if is part of the 5-element subset, and 0 otherwise. The number of pairs of consecutive integers in a 5-element selection is given by the sum . By linearity of expectation, the expected value is equal to the sum of the : To compute , note that the number of selections for which is out of possible selections. Thus ~kingofpineapplz
Solution 2
Alternatively, using linearity of expectation on the chosen subset: there are pairs of integers. There are 29 pairs of consecutive integers out of possible pairs, thus the probability that any given pair is consecutive is . By linearity of expectation, there are such pairs on average.
~MT
Solution 2
We define an outcome as with .
We denote by the sample space. Hence. .
: There is only 1 pair of consecutive integers.
: is the single pair of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Denote , , , , . Hence, is the number of outcomes satisfying
Therefore, .
: is the single pair of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: is the single pair of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: is the single pair of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: There are 2 pairs of consecutive integers.
: and are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: and are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: and are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: and are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: and are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: and are two pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: There are 3 pairs of consecutive integers.
: , and are three pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: , and are three pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: , and are three pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: , and are three pairs of consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Similar to our analysis for Case 1.1, .
: There are 4 pairs of consecutive integers.
In this case, are consecutive integers.
We denote by the collection of outcomes satisfying this condition. Hence, is the number of outcomes satisfying
Hence, .
Therefore, the average number of pairs of consecutive integers is
Therefore, the answer is .
Solution 3 (casework bash)
We will proceed with some casework. Let be the number of sets of consecutive numbers in the subset. Note that the maximum possible value of is Case : There is only one way to arrange the distribution of the number of elements in each block here. There are ways to arrange where that block of numbers will go, so a total of ways for this case. Case : There are ways to arrange the distribution of the number of elements in each block here. (4-1, 3-2, 2-3, 1-4). There are ways to arrange where those two blocks of numbers go among the other numbers such that no two of those blocks are adjacent to each other. Total for this case: Case : By Stars and Bars, there are ways to arrange the distribution of the number of elements in each block here. There are ways to arrange where those three blocks of numbers go among the other numbers such that no two of those blocks are adjacent to each other. Total for this case: Case : By Stars and Bars, there are ways to arrange the distribution of the number of elements in each block here. There are ways to arrange where those four blocks of numbers go among the other numbers such that no two of those blocks are adjacent to each other. Total for this case: Since the last case doesn't contribute to the expected value sum, we can ignore it. Using the expected value formula, our desired value is
-fidgetboss_4000
~Steven Chen (www.professorchenedu.com)
See Also
2021 Fall AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 22 |
Followed by Problem 24 |
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 12 Problems and Solutions |
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