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

(Problem 2)
(Added problem and solution)
Line 1: Line 1:
The 2019 AIME I takes place on March 14, 2019.
 
 
 
==Problem 2==
 
==Problem 2==
 
Jenn randomly chooses a number <math>J</math> from <math>1, 2, 3,\ldots, 19, 20</math>. Bela then randomly chooses a number <math>B</math> from <math>1, 2, 3,\ldots, 19, 20</math> distinct from <math>J</math>. The value of <math>B - J</math> is at least <math>2</math> with a probability that can be expressed in the form <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
 
Jenn randomly chooses a number <math>J</math> from <math>1, 2, 3,\ldots, 19, 20</math>. Bela then randomly chooses a number <math>B</math> from <math>1, 2, 3,\ldots, 19, 20</math> distinct from <math>J</math>. The value of <math>B - J</math> is at least <math>2</math> with a probability that can be expressed in the form <math>\frac{m}{n}</math> where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
  
 
==Solution==
 
==Solution==
Wait next year for it...
+
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} = {1}{10}</math>. Therefore the answer is <math>\frac{9}{20} \rightarrow \boxed{029}</math>.
  
 
==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 19:48, 14 March 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

Realize that 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 $P = \frac{38}{380} = {1}{10}$. Therefore the answer is $\frac{9}{20} \rightarrow \boxed{029}$.

See Also

2019 AIME I (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS