The Grey Labyrinth is a collection of puzzles, riddles, mind games, paradoxes and other intellectually challenging diversions. Related topics: puzzle games, logic puzzles, lateral thinking puzzles, philosophy, mind benders, brain teasers, word problems, conundrums, 3d puzzles, spatial reasoning, intelligence tests, mathematical diversions, paradoxes, physics problems, reasoning, math, science.

Message body

 Emoticons View more Emoticons
Options
HTML is OFF
BBCode is ON
Smilies are ON
 Disable BBCode in this post Disable Smilies in this post

 All times are GMT
 Jump to: Select a forum Puzzles and Games----------------Grey Labyrinth PuzzlesVisitor Submitted PuzzlesVisitor GamesMafia Games Miscellaneous----------------Off-TopicVisitor Submitted NewsScience, Art, and CulturePoll Tournaments Administration----------------Grey Labyrinth NewsFeature Requests / Site Problems
 Topic review
Author Message
mith
Posted: Wed Mar 30, 2011 4:16 pm    Post subject: 1

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.
Lepton*
Posted: Wed Mar 30, 2011 2:49 pm    Post subject: 0

Looks good Zahariel, but your subtraction in the last step is wrong, I think; I find a probability of 1/12.
ralphmerridew
Posted: Mon Mar 28, 2011 10:42 pm    Post subject: -1

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.
Zag
Posted: Mon Mar 28, 2011 9:11 pm    Post subject: -2

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.
Zahariel
Posted: Mon Mar 28, 2011 8:07 pm    Post subject: -3

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!)
Lepton
Posted: Mon Mar 28, 2011 5:39 am    Post subject: -4

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)