Amanufacturing machine has a 10% defect rate. if 4 items are chosen at random, what is the probability that at least one will have a defect?

In probability, there is an equation appropriate for repeated trials. This concept is used to find the probability of an event happening when a number of a series of trials are done. The equation is:

n!/r!(n-r)! * p^(n-r) * q^r, where

n is the number of total number of trials
r is the number of success from the trials
p is the binomial probability of the event happening
q is equal to 1 - p

It is important to note that probability is a part of the whole. In this problem, we are to find the probability of at least one defect. That means it encompasses also the probability of 2, 3, and 4 defects. To be easier, let's just find the probability of no defect, then find subtract this to 1, to get the answer. 

So, r=0 defects, n=4 trials, p = 10% defect rate, q = 90%. Substituting,

4!/0!(4-0)! * (0.1)^(4-0) * (0.9)^(0) = 0.0001

That means the probability of at least one defect is 1 - 0.0001 = 0.9999.

He will need 2,898 pieces of paper