Difference between revisions of "2012 AMC 12B Problems/Problem 13"
(→Problem) |
(→Solution) |
||
Line 5: | Line 5: | ||
<math>\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{25}{36}\qquad\textbf{(C)}\ \frac{5}{6}\qquad\textbf{(D)}\ \frac{31}{36}\qquad\textbf{(E)}\ 1</math> | <math>\textbf{(A)}\ \frac{1}{2}\qquad\textbf{(B)}\ \frac{25}{36}\qquad\textbf{(C)}\ \frac{5}{6}\qquad\textbf{(D)}\ \frac{31}{36}\qquad\textbf{(E)}\ 1</math> | ||
− | == | + | ==Solutions== |
===Solution 1=== | ===Solution 1=== |
Revision as of 22:03, 10 October 2020
Problem
Two parabolas have equations and
, where
and
are integers, each chosen independently by rolling a fair six-sided die. What is the probability that the parabolas will have a least one point in common?
Solutions
Solution 1
Set the two equations equal to each other: . Now remove the x squared and get x's on one side:
. Now factor
:
. If a cannot equal
, then there is always a solution, but if
, a
in
chance, leaving a
out
, always having at least one point in common. And if
, then the only way for that to work, is if
, a
in
chance, however, this can occur
ways, so a
in
chance of this happening. So adding one thirty sixth to
, we get the simplified fraction of
; answer
.
Solution 2
Proceed as above to obtain . The probability that the parabolas have at least 1 point in common is 1 minus the probability that they do not intersect. The equation
has no solution if and only if
and
. The probability that
is
while the probability that
is
. Thus we have
for the probability that the parabolas intersect.
See Also
2012 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 12 |
Followed by Problem 14 |
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.