1. ## 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)?

2. ## 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

3. ## The Following User Says Thank You to rogojel For This Useful Post:

Swayseker (07-22-2015)

4. ## 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.)

5. ## The Following User Says Thank You to Con-Tester For This Useful Post:

Swayseker (07-22-2015)

6. ## Re: Counting problem?

Only 2 billion trials, not enough, try 2 billion and ten!

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