2012 AMC 12B Problems/Problem 13
Contents
[hide]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?
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 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 |
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All AMC 12 Problems and Solutions |