Question

binomial probability distribution ...

Answer 1

in every text ive read: p=success, q = 1-p = failure

Answer 2

how come the setup:\[\binom{6}{3}(\frac{1}{2})^3~(\frac{1}{3})^3\]produces a correct result?

Answer 3

the probability of rolling 3 primes and 3 composites on 6 dice ....

apparently 1 is not prime or composite ...

Answer 4

Personally, I wouldn't consider 1 prime or composite. Prime implies it has exactly 2 divisors, and 1 only has a single divisor. And composite means that it has 2 (not necessarily distinct) prime factors. So that solution makes sense to me, although basing it on the binomial distribution seems a bit unusual.

Answer 5

i took it a different way, but when the solution poped up they had used the binomial setup

Answer 6

3*3*3*2*2*2 = 216, 6!/(3!3!) = 20 ways to permute it

6^6 different outcomes

20(216)/6^6

Answer 7

im not sure if the results using the binomial was a fluke, or if its a valid approach for some different specifics

Answer 8

It should be a valid approach.You have \(\displaystyle \binom63\) different orderings, and \(\displaystyle \left(\frac{1}{2}\right)^3\left(\frac{1}{3}\right)^3\) chance of a given ordering. I could even imagine it being expanded to multinomials for similar problems with more than 2 outcomes possible.

Answer 9

ill have to mull it over for the next few decades before i make a ruling :) thnx tho

Answer 10

You're welcome. And I'm finally a champion :)

Answer 11

lol, i was wondering if they still had achievements like that