Question 1207516: Assume that a procedure yields a binomial distribution with a trial repeated
n = 12 times. Use either the binomial probability formula (or a technology like Excel or StatDisk) to find the probability of k = 9 successes given the robability q = 0.33 of success on a single trial.
(Report answer accurate to 4 decimal places.)
P(X=k)=
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! probability of success is normally p.
q is then the probability of no success.
k is normally x
for your problem, i have:
p = .33
q = 1 - p
n = 12
x = 9
formula is p(x) = p^x * q^(n-x) * c(n,x)
c(n,x) = n! / (x! * (n-x)!)
for your problem, the formula becomes:
p(9) = .33^9 * .67 ^ (12 - 9) * c(12,9) = .0030709486.
in your terminology, i would get:
n = 12
k = 9
q = .33
(1-q) = .67
formula would become:
p(k) = .33^k * .67^(n-k) * c(n,k)
when n = 12 and k = 9 and q = .33, formula becomes:
p(9) = .33^9 * .67^3 * c(12,9).
same result.
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