Difference between revisions of "2019 AIME I Problems/Problem 2"
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==Solution== | ==Solution== | ||
− | + | 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== | ==Solution 2== | ||
− | This problem is | + | This problem is asks 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=\boxed{029}</math>. |
-Fidgetboss_4000 | -Fidgetboss_4000 | ||
==Solution 3== | ==Solution 3== | ||
− | + | Create a grid using graph paper, with 20 columns for the values of J from 1 to 20 and 20 rows for the values of B from 1 to 20. Since <math>B</math> cannot equal <math>J</math>, we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since <math>B - J</math> must be at least <math>2</math>, we can mark the line where <math>B - J = 2</math>. Now we sum the number of squares that are on this line and below it. We get <math>171</math>. Then we find the number of total squares, which is <math>400 - 20 = 380</math>. Finally, we take the ratio <math>\frac{171}{380}</math>, which simplifies to <math>\frac{9}{20}</math>. Our answer is <math>9+20=\boxed{029}</math>. | |
==Solution 4== | ==Solution 4== |
Revision as of 10:46, 26 November 2019
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
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 asks 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
Solution 3
Create a grid using graph paper, with 20 columns for the values of J from 1 to 20 and 20 rows for the values of B from 1 to 20. Since cannot equal , we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since must be at least , we can mark the line where . Now we sum the number of squares that are on this line and below it. We get . Then we find the number of total squares, which is . Finally, we take the ratio , which simplifies to . Our answer is .
Solution 4
We can see that if B chooses 20, J has options 1-18, such that . If B chooses 19, J has choices 1-17. By continuing this pattern, B will choose 3 and J will have 1 option. Summing up the total, we get as the total number of solutions. The total amount of choices is (B and J must choose different numbers), so the probability is . Therefore, the answer is
-eric2020
Video Solution
https://www.youtube.com/watch?v=lh570eu8E0E
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.