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Thread: Counting problem?

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    Counting problem?




    NOW SOLVED. SPECIAL THANKS TO rogojel and con-tester. This is probably a simple problem, but I have been through my 8 counting rules about different arrangements, etc, and cannot find one that applies. If 4% of the population is Christian and 19% are Jewish, what is the probability, when selecting a forum of 9 people, that 3 Christians and 6 Jewish people are chosen? This is assuming random selection (ie every element of the population has an equal chance of being selected). Now I cannot see any obvious equiprobable permutations or equiprobable sets (combinations). For instance a permutation that has more Jews in seems more probable than one with more Christians so I cannot use number of permutations. If I start with probablity of Jew and not (other or Christian)x(probability..)etc... I get a complicated solution that I need the number of permutations for, etc. Does someone know of a good method (not necessarily with an answer)?
    Last edited by Swayseker; 07-24-2015 at 11:42 AM.

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    Re: Counting problem?

    hi,
    This is described by a multinomial distribution - you obviously have at least 3 religions in your population Christian, Jewish and Other. In general the distribution is given by P(Xc,Xj, Xo)= N!/(n1!*n2!*n3!) (p1^n1*p2^n2*p3^n3).

    in your case N=9, n1=3, p1=0.04, n2=6, p2=0.19, n3=0 , p3=0.77 amd you need to apply the formula.

    Regards

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    Swayseker (07-22-2015)

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    Re: Counting problem?

    Naandsę.

    You could also use conditional probability plus binomial distribution to solve this problem as follows:
    P(A∩B) = P(A|B)×P(B) with A = “Select 3 Christians and 6 Jews” and B = “Select only Christians or Jews.”

    (The result I calculate in this way has been verified through Monte Carlo simulation with 2,000,000,000 trials.)

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    Swayseker (07-22-2015)

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    Re: Counting problem?


    Only 2 billion trials, not enough, try 2 billion and ten!
    Stop cowardice, ban guns!

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