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June 1

Ergodic theory

The article mentions nothing about its history like when was it found and who found it? How far we have reached in proving this theory? Basically what is our progress about it?65.128.168.2 (talk) 03:22, 1 June 2012 (UTC)[reply]

Not an answer to your questions, but you may want to check out theory (mathematics). Mathematicians do not "prove" a theory. The word "theory" in math refers to a general body of ideas and theorems, not a "something which may or may not be true and needs to be proved or refuted". Staecker (talk) 12:43, 1 June 2012 (UTC)[reply]

The article does have a little history. It says "The first result in this direction is the Poincaré recurrence theorem". -- Poincaré_recurrence_theorem explains that it was published by Henri_Poincaré in 1890. SemanticMantis (talk) 14:15, 1 June 2012 (UTC)[reply]

Probability

If there is a set that contains 10 members, called set A, which contains {a, b, c, d, e, f, g, x, y, z} and there is a subset of set A that has three members, called set B, containing {x, y, z} and there is another subset of set A called set C, which contains 3 random members of set A, then the chance that one or more members of B are also members of B is 3/10 + 3/9 + 3/8, which equals 108/360 + 120/360 + 125/360. Right? And when you add them up, you get 353/360, which is approxiomately 98 percent. So the chance that a member of set B is a member of set C is almost 98 percent. Is that all right? Legolover26 (talk) 16:08, 1 June 2012 (UTC)[reply]

Does that 98% feel right ? Ball park estimate - the probaility that a random member of A is not in C is about 2/3. So the probability that we choose 3 members of A (maybe with repetitions) and none of them is in C is approx. (2/3)³, which is 8/27 or more than 1/4. That's a long way from 2%. Gandalf61 (talk) 16:31, 1 June 2012 (UTC)[reply]

There are (3!10) = 120 C-sets. (Here (3!10) is the binomial coefficient.) The number of C-sets having no element in common with set B is (3!7) = 35. The number of C-sets having one or more element in common with set B is 120−35 = 85. The chance that one or more members of C are also members of B is 85/120 = 71.833%. Bo Jacoby (talk) 16:44, 1 June 2012 (UTC).[reply]

I believe you. I didn't think that really made sense. It's kind of like that problem where I insist that if you flip 3 coins, the chance that all 3 of them will be on the same side is 1/2, because when you do that, there are always 2 coins that end up on the same side, and the chance that the third coin will be the same is 1/2 exactly. Legolover26 (talk) 17:08, 1 June 2012 (UTC)[reply]

No, if you flip 3 coins there are 8 possible outcomes, all are equally likely, and only two have all 3 coins on the same side, so the probability is 1/4. The side with at least two coins and the side of the remaining coin aren't independent, so you can't multiply the probabilities together like that. Hut 8.5 18:59, 1 June 2012 (UTC)[reply]

In your original proposition, where you say "the chance that one or more members of B are also members of C is 3/10 + 3/9 + 3/8" you are counting some possible outcomes more than once. Your sum should be

P(c₁ in B) = 3/10 = 108/360

P(c₂ in B & c₁ not in B) = (3/9)(252/360) = 84/360

P(c₃ in B & c₁,c₂ not in B) = (3/8)(168/360) = 63/360

So the correct probability is 108/360 + 84/360 + 63/360 = 255/360 or 85/120 as quoted by Bo jacoby. - Sussexonian (talk) 09:05, 3 June 2012 (UTC)[reply]

Probably easier to find the probability it does not happen - 7/10 * 6/9 * 5/8 = 35/120

Therefore the probability it does is: 1 - 35/120 = 85/120 86.184.110.22 (talk) 19:17, 4 June 2012 (UTC)[reply]