Question 1072486
this looks like a binomial probability type problem.


the formula is:


p(x) = p^x * q^(n-x) * c(n,x)


p is the probability that a student is a minority.
this makes p = .33


q is the probability that a student is not a minority.
this makes q = 1 - .33 = .67


c(n,x) is the number of ways you can get x minority students out of n students.


n is the total number of students.


you have n = 8, p = .33 and q = .67


c(n,x) is equal to n! / (x! * (n-x)!)


you want to know the probability that more than 5 out of the 8 students will be a minority.


this means you want to know the probability that 6 or 7 or 8 students out of 8 students is a minority.


using the formula of p(x) = p^x * q^(n-x) * c(n,x), then you want the sum of:


p(6) + p(7) + p(8)


p(6) = .33^6 * .67^2 * c(8,6)
p(7) = .33^7 * .67^1 * c(8,7)
p(8) = .33^8 * .67^0 * c(8,8)


c(8,6) = 8! / (6! * 2!) = 56/2 = 28
c(8,7) = 8! / (7! * 1!) = 8
c(8,8) = 8! / (8! * 0!) = 1


formulas become:


p(6) = .33^6 * .67^2 * 28 = .016233
p(7) = .33^7 * .67^1 * 8 = .00228
p(8) = .33^8 * .67^0 * 1 = .00014


add these up and round to the 4th decimal place and you get:


p(6) + p(7) + p(8) = .0187