Difference between revisions of "2012 AMC 12B Problems/Problem 13"
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<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> | ||
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===Solution 1=== | ===Solution 1=== | ||
Revision as of 21:03, 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. 
