Question 1179724
the probability of getting at lest 3 is equal to 1 minus the probability of getting less than 3.


the probability of getting less than 3 is equal to the probability of getting 0 or 1 or 2 only.


this is a binomial problem where p(x) = c(n,x) * p^x * q^(n-x).
n = 100
x = 0 or 1 or 2
c(n,x) = n! / (x! * (n-x)!
specifically, ...


p(one smartie) = 30/100 = 3/10 = .3
p(anything but one smartie) = 1 - .3 = .7


since you are replacing the chocolate bar each time, the probability remains the same for each draw.


p(0) = .3^0 * .7^(100 - 0) * c(100,0) = 3.23447651 * 10^-16
p(1) = .3^1 * .7^(100 - 1) * c(100,1) = 1.38620422 * 10^-14
p(2) = .3^2 * .7^(100 - 2) * c(100,2) = 2.94073323 * 10^-13


the total probability of getting less than 3 is equal to 3.08258813 * 10^-13.


the total probability of getting at least 3 is 1 minus that.


my calculator couldn't handle it, so i went to excel.


this is what i got.


p(less than 3) = 3.08258813312174 * 10^-13


p(3 or greater) = 1 minus p(less than 3) = 9.99999999999686 * 10^-1


i went a little further.


the best excel could give me is:


p(3 or greater) = 0.999999999999692


i resorted to calculating it manually and the best that i could do is:


p(3 or greater) = 0.999999999999601741186687826


it's kind of like a moot point.


if you round your answer to anything less than 12 digits, the answer is p(3 or greater) = 1.


sticking to the answer that the calculator gave me, i would go with p(3 or greater) = 1 or p(3 or greater) = 1 minus p (less than 3) = 1 minus 3.08258813 * 10^-13.