Skip to main content

2012 AMC 12B Problem 13

Problem 13 of 25IntermediateCounting & Probability

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?

Answer choices

Show solution

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 bd.b\neq d. The probability that a=ca=c is 16,\tfrac16, and the probability that bdb\neq d is 56,\tfrac56, so the probability of no common point is 1656=536.\tfrac16\cdot\tfrac56=\tfrac5{36}. The probability of at least one common point is 1536=3136.1-\tfrac5{36}=\tfrac{31}{36}. Thus, the correct answer is D.

More practice

Concepts: basic probability · complementary counting · dice (probability)

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.