The lenght of a bar, X has a distribution N(10,2^2). The bars are separeted in 3 categories A B and C A for lenght less than 8 B for lenght between 8 and 12 and C for lenght greater than 12

what is the probability that if we choose 15 bars we have the same number of bars of category A B and C.

Well tried to solve it that way:

we have to have 5 bars in each category.

And i found that the probability of belonging to category A and C is 0.1587 and for B 1-2*0.1587 I think I'm right assuming that.

then I think the probability we seek is given by:

Cn(15,5)*(0.1587^10)*(1-2*0.1587)^5

I can't see why it's not right but the truth is that probably it is wrong since I don't get the result I should.

Can anyone see where Is my mistake? perhaps I shouldn't have used Combinations and maybe permutations or something...but cant see why...can anyone explain me?

Thanks for your time