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2012 AMC 12B Problem 13

Problem 13 of 25IntermediateCombinatoricsProbability & Statistics

Two parabolas have equations y=x2+ax+by = x^2 + ax + b and y=x2+cx+d,y = x^2 + cx + d, where a,a, b,b, c,c, and dd are integers (not necessarily different), each chosen independently by rolling a fair six-sided die. What is the probability that the parabolas have at least one point in common?

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Solution

The parabolas meet where x2+ax+b=x2+cx+d,x^2+ax+b=x^2+cx+d, i.e. ax+b=cx+d.ax+b=cx+d. This has no solution exactly when the lines are parallel and distinct: a=ca=c and b≠d.b\neq d. The probability that a=ca=c is 16,\tfrac16, and the probability that b≠db\neq d is 56,\tfrac56, so the probability of no common point is 16⋅56=536.\tfrac16\cdot\tfrac56=\tfrac5{36}. The probability of at least one common point is 1−536=3136.1-\tfrac5{36}=\tfrac{31}{36}. Thus, the correct answer is D.
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Tagged: basic probability · complementary counting · dice (probability)

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