Answer:
31.89%
Step-by-step explanation:
This situation can be modeled with the Negative Binomial Distribution, where <em>the probability of having r “failures” before k “successes” occur is given by</em>
![\large P(X=k)=\binom{k+r-1}{k}(1-p)^rp^k](https://tex.z-dn.net/?f=%5Clarge%20P%28X%3Dk%29%3D%5Cbinom%7Bk%2Br-1%7D%7Bk%7D%281-p%29%5Erp%5Ek)
<em>being p the probability that a “success” occurs.</em>
is the number of combinations of m elements taken n at a time.
In the specific case of this problem we have “success” is having a copy with a defect, with probability 0.1, k=1 and r=6 (6 “failures” before 1 “success”).
Computing the formula either by hand or with a computer, we get
![\large P(X=1)=\binom{1+6-1}{1}(1-0.1)^60.1=6*0.9^6*0.1=0.31886\approx 31.89\%](https://tex.z-dn.net/?f=%5Clarge%20P%28X%3D1%29%3D%5Cbinom%7B1%2B6-1%7D%7B1%7D%281-0.1%29%5E60.1%3D6%2A0.9%5E6%2A0.1%3D0.31886%5Capprox%2031.89%5C%25)