Alex
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 Apr29 comment How do you calculate the probability of simultaneous events? Let A1, A2, A3, A4 be the events. Let P(A) be the probability of the event. So P(A1) = P(A2) = P(A3) = P(A4) = 0.1. You are searching for the probability of the union. P(A1 union A2 union A3 union A4) = P(A1) + P(A2) + P(A3) + P(A4) - P(A1)P(A2) - P(A1)P(A3) - P(A1)P(A4) - P(A2)P(A3) - P(A2)P(A4) - P(A3)P(A4) + P(A1)P(A2)P(A3) + P(A1)P(A2)P(A4) + P(A1)P(A3)P(A4) + P(A2)P(A3)P(A4) - P(A1)P(A2)P(A3)P(A4). Do the calculation and you'll have the exact probability. Hope this helps. Mar31 comment When two voters meet, they switch allegiance; might they all ally with the same candidate? No, 'always' is not implied. The question is about whether even one possibility exists. Feel free to edit it if you feel appropriate. Jan5 comment Easiest way to prove that $2^{\aleph_0} = c$ Your answer is good. I understand what the theorem states, I looked it up before posting this. I'll wait a while to see if anybody comes up with a simpler solution, and if that doesn't happen I'll accept the answer. Forget an explicit bijection, but could you possibly explicitly suggest examples for the desired injections? Thanks Jan5 comment Easiest way to prove that $2^{\aleph_0} = c$ @isomorphismes I don't need to prove that c = aleph 1. That would require the Continuum hypothesis and further complicate things. Also, I am looking for the simplest way, your edit might suggest that I am looking for the original proof. Jan4 comment How to prove some properties of partitions of finite subsets of N? @gnometorule Yes, I made a mistake. I corrected it to provide a non-recursive definition for the progression. Jan4 comment How to prove some properties of partitions of finite subsets of N? @Ilya I don't see what is your point, could you explain a bit more? Jan4 comment How to prove some properties of partitions of finite subsets of N? @Amr That is not a counterexample. In the second set, 2, 5, 8 are elements of an arithmetic progression, and also 5, 6, 7, 8, 9 are elements of another arithmetic progression. Nov15 comment Four men seated in a boat puzzle Yes, that lends an idea of the process. Thanks again, it's all fine now. Nov15 comment Four men seated in a boat puzzle Alright, that makes sense. Thanks. Nov15 comment Four men seated in a boat puzzle It would be most helpful if you could edit your post to contain that additional labeling of conditions and establishing those relationships as much as that could be done. If you could do that, than that would be the perfect solution to this question. Nov15 comment Four men seated in a boat puzzle Good answer. I'm just interested if we could use some logic statements, or logic functions with predicates to mathematically write out all of this. Of course, we could write those requests for each man as logical functions using predicates and establish the conditions, but besides that, is there any way to elegantly and mathematically write out all that you said, and the fact that only "BACD" is the solution to the exercise?