Assume that a box contains 3 defective widgets and 8 acceptable widgets. There are 3 inspectors, each of whom selects a widget from the box at random, inspects it, and then replaces it in the box. Each inspector makes his or her inspection at a different time.
What is the probability that at least one inspector will find a defective widget?
Currently I have an answer of (819/1331), however it's wrong. I got this answer by making the Pr[success] = choosing defective widget, p = (3/11), and n = 3.
Therefore, I did 1 – Pr[0 success] –> 1 – (3C0 * (3/11)^0 * (8/11)^3)
Can anyone explain why it's wrong and how to get the right answer?
Best Answer
The probability for at least one success among the three tries is indeed: $1-{}^3\mathrm C_0 {(\frac{3}{11})}^{0}{(\frac{8}{11})}^{3} = \dfrac{819}{1331}$
What makes you think it is not? You are correctly modelling the count of defective widgets as a Binomial Random Variable with success rate $3/11$ and trial amount $3$.