Counting problem?

#1
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)?
 
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rogojel

TS Contributor
#2
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
Naandsê.

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.)