Wikipedia:Reference desk/Archives/Mathematics/2012 November 2
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November 2
[edit]Generalizing the birthday problem
[edit]Can anyone find a good way to calculate the following: you have 15 people, and each person has 5 distinct birthdays, chosen uniformly from a 50 day year. What is the probability that there is a day on which at least 5 people have a birthday?--149.148.254.159 (talk) 14:28, 2 November 2012 (UTC)
- Forgot to add: each person's birthdays are chosen independently of other people's birthdays.--149.148.254.159 (talk) 14:29, 2 November 2012 (UTC)
You can roughly approximate this as follows. Each day has on average birthdays. If we approximate the number of birthdays per day as a Poisson process, then the probablity that a day has less than 5 birthdays is:
The probability that all days have less than 5 birthdays is p^50, the probability that one or more days have 5 or more birthdays is thus 1 - p^50 = 0.6084. Count Iblis (talk) 17:12, 2 November 2012 (UTC)
- OP here. It's not hard to calculate p exactly using binomial distributions, but when you then calculate p^50, you seem to be treating days as independent. Are you sure that's justified?--80.109.106.49 (talk) 18:17, 2 November 2012 (UTC)
- can a person have all their birthdays on the same day or do they have to be distinct? 122.60.245.210 (talk) 08:46, 3 November 2012 (UTC)
- All must be distinct.--80.109.106.49 (talk) 17:39, 3 November 2012 (UTC)
- OK then, we can use direct numerical simulation, here with the R programming language:
- All must be distinct.--80.109.106.49 (talk) 17:39, 3 November 2012 (UTC)
> table(replicate(1e6,any(rowSums(replicate(15,sample(c(rep(1,5),rep(0,45)))))==5)))
FALSE TRUE 576416 423584
So about 42.3% of the time. HTH, Robinh (talk) 08:11, 4 November 2012 (UTC)
- This answer is surprising to me, since assuming days are independent, and calculating p exactly, I get 47.3%. Of course they're not independent, but rather knowing one day had 5 birthdays decreases the probability that another day did; equivalently, knowing one day did not have 5 birthdays increases the probability that another day did. So the actual answer should be greater than 47.3%, yet your number is smaller.--80.109.106.49 (talk) 10:06, 4 November 2012 (UTC)
- Apologies, I was checking for *exactly* five birthdays and you asked for *at least* five:
- This answer is surprising to me, since assuming days are independent, and calculating p exactly, I get 47.3%. Of course they're not independent, but rather knowing one day had 5 birthdays decreases the probability that another day did; equivalently, knowing one day did not have 5 birthdays increases the probability that another day did. So the actual answer should be greater than 47.3%, yet your number is smaller.--80.109.106.49 (talk) 10:06, 4 November 2012 (UTC)
R> table(replicate(1e6,any(rowSums(replicate(15,sample(c(rep(1,5),rep(0,45))))) >= 5)))
FALSE TRUE 504936 495064
- So I get 49.5%, more or less. Let me know if you want a more accurate value. Sorry about that. Robinh (talk) 21:47, 4 November 2012 (UTC)
- Ah, makes sense. That's plenty accurate, this was just for settling a little debate (which I've reframed; it wasn't actually about people with multiple birthdays). Thanks.--80.109.106.49 (talk) 00:18, 5 November 2012 (UTC)
- So I get 49.5%, more or less. Let me know if you want a more accurate value. Sorry about that. Robinh (talk) 21:47, 4 November 2012 (UTC)
Integration
[edit]I was asked solve the following:
At first, I thought the integration can be evaluated directly as,
Then I had to look for the value of , but found things complicated. When I turn to basics, evaluating this integral,
- ,
I assume, it should be , however, wolframalpha ignores the constant. Am I wrong somewhere?--Almuhammedi (talk) 14:53, 2 November 2012 (UTC)
- This problem is ill-defined as stated. The meaning of "solve" needs to be clarified, and we need to know what is a function of what. Looie496 (talk) 17:27, 2 November 2012 (UTC)
- I'm not sure of the proper descriptive term (you may say, evaluate).--Almuhammedi (talk) 18:20, 2 November 2012 (UTC)
- Urgh. Wolframalpha appears to interpret this as the definite integral from x=0 to x=x0, and that it is to find x0. More obvious if you try this. Ignore its result. — Quondum 16:17, 5 November 2012 (UTC)
History of knowledge of various geometric shapes
[edit]Our articles circle, hyperbola, and parabola have history sections. But ellipse, triangle, Reuleaux triangle, curve of constant width, quadrilateral, parallelogram, and others do not. Can someone give references on when each of these was first studied, and what was known at that time? Also, I invite any interested editors to contribute history sections to these articles. Duoduoduo (talk) 16:50, 2 November 2012 (UTC)
- Franz Reuleaux lived from 1829 to 1905, so that is some indication for his triangle. Euclid's Elements was one of the first works on geometry and contains some basic facts on triangles. Rojomoke (talk) 18:03, 2 November 2012 (UTC)
- Apollonius of Perga did some quite remarkable work on conics, including ellipses, he studied their normals and centers of curvature. See [1] Aristaeus the Elder did some earlier work as did Euclid.--Salix (talk): 18:15, 2 November 2012 (UTC)
- Some of those were likely studied "in antiquity". That is, before writing. Therefore, we don't know the specifics. For example, the hypotenuse of a right triangle being longer than either leg but less than the total of the two legs was probably known way back, although the Pythagorean theorem wasn't discovered until later (when they knew how to find irrational square roots). StuRat (talk) 05:06, 4 November 2012 (UTC)