Question 1120617
probability of failure is .02
probability of no failure is 1 - .02 = .98


the formula to use is p(x) = p^x * q^x * c(n,x)


p is the probability of failure
q is the probability of no failure
x is the number of failures.
n-x is the number of no failures
c(n,x) is the number of ways to get sets of x elements out of sets of n elements without worrying about order within each set.


total probability is always equal to 1.


probability of at least 1 failure is equal to 1 minus probability of no failures.


in your problem.


n = 8
p = .02
q = .98
x = 0 through 8.


p(0) = .02^0 * .98^(8-0) * c(8,0)


this becomes p(0) = 1 * .98^8 * c(8,0) which becomes:


p(0) = 1 * .850763023 * 1 which becomes:


p(0) = 850763023.


probability of at least 1 failure is therefore equal to 1 minus .850763023.


this results in probability of at least 1 failure = .149236977.


all the probabilities of failure are captured in the following excel spreadsheet.


<img src = "http://theo.x10hosting.com/2018/072901.jpg" alt="$$$" >


n! = n * (n-1) * (n-2) * (n-3) * ..... * (n-n + 1)


for example, if n = 5, then n! = 5 * 4 * 3 * 2 * 1


n is 5
n-1 is 4
n-2 is 3
n-3 is 2
n-4 is 1 
n-4 is the same as n-n+1 = 5-5+1 = 0+1 = 1