Difference between revisions of "2019 AIME I Problems/Problem 2"
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Realize that by symmetry, the desired probability is equal to the probability that <math>J - B</math> is at most <math>-2</math>, which is <math>\frac{1-P}{2}</math> where <math>P</math> is the probability that <math>B</math> and <math>J</math> differ by 1 (no zero, because the two numbers are distinct). There are <math>20 * 19 = 380</math> total possible combinations of <math>B</math> and <math>J</math>, and <math>1 + 18 * 2 + 1 = 38</math> ones that form <math>P</math>, so <math>P = \frac{38}{380} = \frac{1}{10}</math>. Therefore the answer is <math>\frac{9}{20} \rightarrow \boxed{029}</math>. | Realize that by symmetry, the desired probability is equal to the probability that <math>J - B</math> is at most <math>-2</math>, which is <math>\frac{1-P}{2}</math> where <math>P</math> is the probability that <math>B</math> and <math>J</math> differ by 1 (no zero, because the two numbers are distinct). There are <math>20 * 19 = 380</math> total possible combinations of <math>B</math> and <math>J</math>, and <math>1 + 18 * 2 + 1 = 38</math> ones that form <math>P</math>, so <math>P = \frac{38}{380} = \frac{1}{10}</math>. Therefore the answer is <math>\frac{9}{20} \rightarrow \boxed{029}</math>. | ||
+ | ==Solution 2== | ||
+ | This problem is basically asking how many ways there are to choose 2 distinct elements from a 20 element set such that no 2 elements are adjacent. Using the well-known formula <math>\dbinom{n-k+1}{k}</math>, there are <math>\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171</math> ways. Dividing 171 by 380, our desired probability is <math>\frac{171}{380} = \frac{9}{20}</math>. Thus, our answer is <math>9+20=29</math>. | ||
+ | -Fidgetboss_4000 | ||
==See Also== | ==See Also== | ||
{{AIME box|year=2019|n=I|num-b=1|num-a=3}} | {{AIME box|year=2019|n=I|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 21:30, 14 March 2019
Contents
Problem 2
Jenn randomly chooses a number from . Bela then randomly chooses a number from distinct from . The value of is at least with a probability that can be expressed in the form where and are relatively prime positive integers. Find .
Solution
Realize that by symmetry, the desired probability is equal to the probability that is at most , which is where is the probability that and differ by 1 (no zero, because the two numbers are distinct). There are total possible combinations of and , and ones that form , so . Therefore the answer is .
Solution 2
This problem is basically asking how many ways there are to choose 2 distinct elements from a 20 element set such that no 2 elements are adjacent. Using the well-known formula , there are ways. Dividing 171 by 380, our desired probability is . Thus, our answer is . -Fidgetboss_4000
See Also
2019 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.