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

Revision as of 03:26, 31 December 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 essentially 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 $\dbinom{n-k+1}{k}$ , 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 $20$ , $J$ can choose from $1$ through $18$ 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 6: / Line 6: @@
 ==Solution 2==
-This problem is asks how many ways there are to choose <math>2</math> distinct elements from a <math>20</math> element set such that no <math>2</math> 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 <math>171</math> by <math>380</math>, 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 essentially asking how many ways there are to choose <math>2</math> distinct elements from a <math>20</math> element set such that no <math>2</math> 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 <math>171</math> by <math>380</math>, our desired probability is <math>\frac{171}{380} = \frac{9}{20}</math>. Thus, our answer is <math>9+20=\boxed{029}</math>.
 -Fidgetboss_4000

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