# 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.

Author Message
Lepton
1:41+ Arse Scratcher

 Posted: Mon Mar 28, 2011 5:39 am    Post subject: 1 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
Daedalian Member

 Posted: Mon Mar 28, 2011 8:07 pm    Post subject: 2 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!)Last edited by Zahariel on Wed Mar 30, 2011 5:49 pm; edited 1 time in total
Zag
Unintentionally offensive old coot

 Posted: Mon Mar 28, 2011 9:11 pm    Post subject: 3 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
Daedalian Member

 Posted: Mon Mar 28, 2011 10:42 pm    Post subject: 4 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*
Guest

 Posted: Wed Mar 30, 2011 2:49 pm    Post subject: 5 Looks good Zahariel, but your subtraction in the last step is wrong, I think; I find a probability of 1/12.
mith
Pitbull of Truth

 Posted: Wed Mar 30, 2011 4:16 pm    Post subject: 6 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.
 Display posts from previous: All Posts1 Day7 Days2 Weeks1 Month3 Months6 Months1 Year by All usersAnonymousLeptonmithralphmerridewZagZahariel Oldest FirstNewest First
 All times are GMT Page 1 of 1

 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
You cannot post new topics in this forum
You can reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum