Question 852124
close but no cigar.
if you multiply that by the number of possible ways you can get 3 things out of 8 things, then you would be right.
this, i believe, is a binomial probability type problem.
the formula for binomial probability is:
p(x) = nCx * p^x * q^(n-x)
in your problem:
n = 8
x = 3
n-x = 5
p = (1/6)
q = (5/6)
using these values, the formula becomes:
p(3) = 8C3 * (1/6)^3 * (5/6)^5
8C3 is equal to (8*7*6) / (1*2*3) which is equal to 56.
(1/6)^3 is equal to .004630
(5/6)^5 is equal to .401878
.004630 * .401878 is equal to .001860
formula becomes:
p(3) = 56 * .004630 * .401878 which is equal to 56 * .001860 which is equal to .104199


you were close.
you calculated correctly for one of the possible ways you can get 3 aces out of 8 serves.
there were 55 more ways to do that.
i'll just list the first 3 so you can see what i mean.
let a = probability to get an ace which is equal to 1/6
let b = probability to not get an ace which is equal to 5/6.
you showed one way it could happen:
aaabbbbb
another way it could happen would be:
aababbbb
another way it could happen would be:
aabbabbb
etc.
so the probability you calculated is correct but for only one of the ways it could happen.
there were 55 other ways it could happen as well.


the binomial probability formula takes that into account.
p(x) = nCx * p^x * q^(*n-x)
nCx is the number of possible ways to get x things from n.
p = probability of the event occurring.
q k= probability of the event not occurring = 1 - p.