Wikipedia:Reference desk/Archives/Mathematics/2010 December 16

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December 16

Probability

I did a puzzle recently that went like: 2010 points are distributed evenly on a circle. If three points are randomly selected, what is the probability that they form a right triangle? I reasoned that the converse of the Mean Value Theorem implies that on a circle there are chords with every possible slope. Therefore the first two points chosen do not matter, only the last point. However, there are two points on a circle that would form a chord at a right angle with the first; thus I answered 1/1004 (or 2/2008). However the answer the book had was 3/2009. Where was my reasoning faulty? 24.92.70.160 (talk) 00:41, 16 December 2010 (UTC)

First note that there are ${\tbinom {2010}{3}}\,\!$ ways to choose 3 points on the circle. Recall that a right triangle inscribed in a circle has a diameter for its hypotenuse (Thales' theorem). So, when choosing any given point, we must also choose the point diametrically opposite if we wish to form a right triangle. After that we can select any of the 2008 other points. Then there are $2010(2008)/2$ ways to form a right triangle from the 2010 points, where the division by two prevents double-counting. The probability is therefore

${\frac {2010(2008)/2}{\tbinom {2010}{3}}}={\frac {3}{2009}}.$

Hope that helps. —Anonymous Dissident^Talk 01:15, 16 December 2010 (UTC)

Your error was forgetting that if the first two points are opposite (which has probability 1/2009) then any third point will work. The total probability becomes: (1/2009 × 1) + (2008/2009 × 2/2008) = 3/2009. PrimeHunter (talk) 02:16, 16 December 2010 (UTC)

Another (equivalent) way of looking at it is: Choose any point. the second must be its antipode, which has probability $1/2009$, or the third must be, which has probability ${\frac {2008}{2009}}{\frac {1}{2008}}$ (where the first fraction is to ensure the second point is not opposite the first point), or the second must be opposite the third, which has probability ${\frac {2008}{2009}}{\frac {1}{2008}}$ (see preceding parenthetical remark). Adding these you get $3/2009$.—msh210℠ 17:03, 16 December 2010 (UTC)

Calculus textbook question

After reading Spivak's Calculus, what is a good continuation into more advanced calculus (I'm looking for a book)? It should be no more rigorous than Spivak, and preferably slightly less rigorous. I am especially interested in partial derivatives and contour/path/line integrals. 24.92.70.160 (talk) 03:38, 16 December 2010 (UTC)

Options abound. There's Folland's Advanced Calculus, or Marsden and Tromba's Vector Calculus, both of which are pretty standard. I've heard very good things about Apostol's Calculus and Linear Algebra (the latter part of volume I and most of volume II should serve your need), although I've never read that one. I also recommend Div, Grad, Curl and all that as a very nice intuitive supplement. If you found Spivak's Calculus too rigorous for comfort, I would advise you to avoid Calculus on Manifolds (also by Spivak), or Rudin's Principles of Mathematical Analysis. Cheers, Ray^Talk 06:15, 16 December 2010 (UTC)

About the title 'Another math question'. This is the maths reference desk so I would guess it was a maths question and since you started another topic I would guess it was 'another'. Please try and think of a better title when submitting questions. Dmcq (talk) 11:55, 16 December 2010 (UTC)

Probability and combinatorics

Hello. I'm looking for good formal introductions to probability and combinatorics – material that might be covered in an undergraduate course, for example. The topics are related but deserve separate treatment, of course, so I'm not looking for a single book for both. Thanks in advance for any recommendations. —Anonymous Dissident^Talk 11:04, 16 December 2010 (UTC)

Did you study our articles on probability and combinatorics? They have references to books. If you need something specific, ask again. Bo Jacoby (talk) 11:40, 18 December 2010 (UTC).