Answer:
P(X>17) = 0.979
Step-by-step explanation:
Probability that a camera is defective, p = 3% = 3/100 = 0.03
20 cameras were randomly selected.i.e sample size, n = 20
Probability that a camera is working, q = 1 - p = 1 - 0.03 = 0.97
Probability that more than 17 cameras are working P ( X > 17)
This is a binomial distribution P(X = r) ![nCr q^{r} p^{n-r}](https://tex.z-dn.net/?f=nCr%20q%5E%7Br%7D%20p%5E%7Bn-r%7D)
![nCr = \frac{n!}{(n-r)!r!}](https://tex.z-dn.net/?f=nCr%20%3D%20%5Cfrac%7Bn%21%7D%7B%28n-r%29%21r%21%7D)
P(X>17) = P(X=18) + P(X=19) + P(X=20)
P(X=18) = ![20C18 * 0.97^{18} * 0.03^{20-18}](https://tex.z-dn.net/?f=20C18%20%2A%200.97%5E%7B18%7D%20%2A%200.03%5E%7B20-18%7D)
P(X=18) = ![20C18 * 0.97^{18} * 0.03^{2}](https://tex.z-dn.net/?f=20C18%20%2A%200.97%5E%7B18%7D%20%2A%200.03%5E%7B2%7D)
P(X=18) = 0.0988
P(X=19) = ![20C19 * 0.97^{19} * 0.03^{20-19}](https://tex.z-dn.net/?f=20C19%20%2A%200.97%5E%7B19%7D%20%2A%200.03%5E%7B20-19%7D)
P(X=19) = ![20C19 * 0.97^{19} * 0.03^{1}](https://tex.z-dn.net/?f=20C19%20%2A%200.97%5E%7B19%7D%20%2A%200.03%5E%7B1%7D)
P(X=19) = 0.3364
P(X=20) = ![20C20 * 0.97^{20} * 0.03^{20-20}](https://tex.z-dn.net/?f=20C20%20%2A%200.97%5E%7B20%7D%20%2A%200.03%5E%7B20-20%7D)
P(X=20) = ![20C20 * 0.97^{20} * 0.03^{0}](https://tex.z-dn.net/?f=20C20%20%2A%200.97%5E%7B20%7D%20%2A%200.03%5E%7B0%7D)
P(X=20) = 0.5438
P(X>17) = 0.0988 + 0.3364 + 0.5438
P(X>17) = 0.979
The probability that there are more than 17 working cameras should be 0.979 for the company to accept the whole batch