An easier probability question on [0,1]

[Lepton]

I don't know how to solve Laramie's problem, so I thought I'd share something sort of similar.

Consider the quadratic f(x) = x^2 + ax + b. a and b are random numbers between 0 and 1. What is the probability that f(x) will have at least one real root?

(note: this is a high-school level problem, so the precise definitions of "random", "probability", and so forth are naively straightforward... but I could post a more formal translation of this, if there is need for it)

[Zahariel]

f(x) has at least one real root if a^2 >= 4b. So we set up an integral: \int_{0}^{1/4} P(a^2 >= 4b) db = 1/4 - \int_{0}^{1/4) P(a^2 < 4b) db = 1/4 - \int_{0}^{1/4} P(a < 2 sqrt (b)) db = 1/4 - 2 \int_{0}^{1/4} sqrt(b) db = 1/4 - 2(1/12) = 1/12. Unless I did the calculus wrong, which is 100% possible as I'm not really much good at calculus.

(edit: I'm bad at math, thanks Lepton!)

[Zag]

Hah! I originally misread this problem as "probability that f(x) will have at least one rational root?" This is true only if SQR( a^2 - 4b ) is rational, and I had no idea how to calculate the chances of that.

Zahariel, your false modesty ill becomes you.

[ralphmerridew]

Probability of at least one rational root is 0.

For any rational x, the set of ordered pairs (a,b) such that x^2+ax+b will be a line segment in [0,1]x[0,1] (has zero measure in that space). The rationals are countable, and the intersection of a countable number of sets with zero measure has zero measure.

[Lepton*]

Looks good Zahariel, but your subtraction in the last step is wrong, I think; I find a probability of 1/12.

[mith]

Yeah, I get 1/12 as well.

Integrating with a as the independent variable cleans things up: The probability of a^2 <= 4b is the area under the curve b = a^2/4 on [0,1], which is 1^3/12 - 0^3/12.