Difference between revisions of "2019 AIME I Problems/Problem 2"
Expilncalc (talk | contribs) (Added problem and solution) |
m (→Solution) |
||
Line 3: | Line 3: | ||
==Solution== | ==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} = {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>. |
==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 21:12, 14 March 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
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
.
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.