Difference between revisions of "2019 AIME I Problems/Problem 2"

Revision as of 10:46, 26 November 2019

Problem 2

Jenn randomly chooses a number $J$ from $1, 2, 3,\ldots, 19, 20$ . Bela then randomly chooses a number $B$ from $1, 2, 3,\ldots, 19, 20$ distinct from $J$ . The value of $B - J$ is at least $2$ with a probability that can be expressed in the form $\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

By symmetry, the desired probability is equal to the probability that $J - B$ is at most $-2$ , which is $\frac{1-P}{2}$ where $P$ is the probability that $B$ and $J$ differ by 1 (no zero, because the two numbers are distinct). There are $20 * 19 = 380$ total possible combinations of $B$ and $J$ , and $1 + 18 * 2 + 1 = 38$ ones that form $P$ , so $P = \frac{38}{380} = \frac{1}{10}$ . Therefore the answer is $\frac{9}{20} \rightarrow \boxed{029}$ .

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 $\dbinom{n-k+1}{k}$ , there are $\dbinom{20-2+1}{2} = \dbinom{19}{2} = 171$ ways. Dividing 171 by 380, our desired probability is $\frac{171}{380} = \frac{9}{20}$ . Thus, our answer is $9+20=\boxed{029}$ . -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 $B$ cannot equal $J$ , we cross out the diagonal line from the first column of the first row to the twentieth column of the last row. Now, since $B - J$ must be at least $2$ , we can mark the line where $B - J = 2$ . Now we sum the number of squares that are on this line and below it. We get $171$ . Then we find the number of total squares, which is $400 - 20 = 380$ . Finally, we take the ratio $\frac{171}{380}$ , which simplifies to $\frac{9}{20}$ . Our answer is $9+20=\boxed{029}$ .

Solution 4

We can see that if B chooses 20, J has options 1-18, such that $B-J\geq 2$ . 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 $18+17+\cdots+1$ as the total number of solutions. The total amount of choices is $20\times19$ (B and J must choose different numbers), so the probability is $\frac{18*19/2}{20*19}=\frac{9}{20}$ . Therefore, the answer is $9+20=\boxed{029}$

-eric2020

Video Solution

https://www.youtube.com/watch?v=lh570eu8E0E

@@ Line 3: / Line 3: @@
 ==Solution==
-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>.
+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=\boxed{029}</math>.
+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
 ==Solution 3==
-Simply 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 we know that <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>.
+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==

2019 AIME I (Problems • Answer Key • Resources)
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