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Difference between revisions of "2012 AMC 12B Problems/Problem 13"

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Proceed as above to obtain <math>x(a-c)=d-b</math>. The probability that the parabolas have at least 1 point in common is 1 minus the probability that they do not intersect. The equation <math>x(a-c)=d-b</math> has no solution if and only if <math>a=c</math> and <math>d\neq b</math>. The probability that <math>a=c</math> is <math>\frac{1}{6}</math> while the probability that <math>d\neq b</math> is <math>\frac{5}{6}</math>. Thus we have <math>1-\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)=\frac{31}{36}</math> for the probability that the parabolas intersect.
 
Proceed as above to obtain <math>x(a-c)=d-b</math>. The probability that the parabolas have at least 1 point in common is 1 minus the probability that they do not intersect. The equation <math>x(a-c)=d-b</math> has no solution if and only if <math>a=c</math> and <math>d\neq b</math>. The probability that <math>a=c</math> is <math>\frac{1}{6}</math> while the probability that <math>d\neq b</math> is <math>\frac{5}{6}</math>. Thus we have <math>1-\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)=\frac{31}{36}</math> for the probability that the parabolas intersect.
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== See Also ==
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{{AMC12 box|year=2012|ab=B|num-b=12|num-a=14}}

Revision as of 22:20, 12 January 2013

Problem

Two parabolas have equations $y= x^2 + ax +b$ and $y= x^2 + cx +d$, where $a, b, c,$ and $d$ 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: $x^2 + ax + b = x^2 + cx + d$. Now remove the x squared and get x's on one side: $ax-cx=d-b$. Now factor $x$: $x(a-c)=d-b$. If a cannot equal $c$, then there is always a solution, but if $a=c$, a $1$ in $6$ chance, leaving a $1080$ out $1296$, always having at least one point in common. And if $a=c$, then the only way for that to work, is if $d=b$, a $1$ in $36$ chance, however, this can occur $6$ ways, so a $1$ in $6$ chance of this happening. So adding one sixth to $\frac{1080}{1296}$, we get the simplified fraction of $\frac{31}{36}$; answer $(D)$.

Solution 2

Proceed as above to obtain $x(a-c)=d-b$. The probability that the parabolas have at least 1 point in common is 1 minus the probability that they do not intersect. The equation $x(a-c)=d-b$ has no solution if and only if $a=c$ and $d\neq b$. The probability that $a=c$ is $\frac{1}{6}$ while the probability that $d\neq b$ is $\frac{5}{6}$. Thus we have $1-\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)=\frac{31}{36}$ 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
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