Difference between revisions of "2011 AMC 12A Problems/Problem 14"
(→Video Solution) |
|||
| Line 24: | Line 24: | ||
https://www.youtube.com/watch?v=u23iWcqbJlE | https://www.youtube.com/watch?v=u23iWcqbJlE | ||
~Shreyas S | ~Shreyas S | ||
| + | |||
| + | this links to problem 11... | ||
== See also == | == See also == | ||
{{AMC12 box|year=2011|num-b=13|num-a=15|ab=A}} | {{AMC12 box|year=2011|num-b=13|num-a=15|ab=A}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 22:06, 14 April 2022
Contents
Problem
Suppose 
and 
are single-digit positive integers chosen independently and at random. What is the probability that the point 
lies above the parabola 
?
Solution
If lies above the parabola, then must be greater than 
. We thus get the inequality 
. Solving this for gives us 
. Now note that 
constantly increases when is positive. Then since this expression is greater than 
when 
, we can deduce that must be less than 
in order for the inequality to hold, since otherwise would be greater than and not a single-digit integer. The only possibilities for are thus 
, 
, and 
.
For 
, we get 
for our inequality, and thus can be any integer from to .
For 
, we get 
for our inequality, and thus can be any integer from to .
For 
, we get 
for our inequality, and thus can be any integer from 
to .
Finally, if we total up all the possibilities we see there are 
points that satisfy the condition, out of 
total points. The probability of picking a point that lies above the parabola is thus 
Video Solution
https://www.youtube.com/watch?v=u23iWcqbJlE ~Shreyas S
this links to problem 11...
See also
| 2011 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 13 |
Followed by Problem 15 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
