Question

3.49. if a lot of size 100 contains 5 defective or more items, the customer will reject it. the customer bases this decision on a sampling process. a sample of size 10 is taken from the lot and if 2 or more of the items in the sample are defective, the lot is rejected. what is the probability that a lot will be rejected if it contains exactly 5 defective items? what is the probability that a lot will be rejected if it contains exactly 3 defective items? what do you think of the customers decision rule? 3.50. the time to failure of a product is well described by the following probability distribution:

Explanation:

Step1: Calculate the total number of ways to choose a sample of size 10 from 100 items

The number of combinations of choosing $n$ items from $m$ items is given by the formula $C(m,n)=\frac{m!}{n!(m - n)!}$. Here, $m = 100$ and $n=10$, so the total number of samples of size 10 is $C(100,10)=\frac{100!}{10!(100 - 10)!}$.

Step2: Calculate the number of non - rejected samples

The lot is rejected if 2 or more of the items in the sample are defective. First, find the number of samples with 0 defective items and 1 defective item.
The number of non - defective items is $100 - 5=95$.
The number of ways to choose 0 defective items (i.e., all 10 from 95 non - defective items) is $C(95,10)=\frac{95!}{10!(95 - 10)!}$.
The number of ways to choose 1 defective item (1 from 5 defective and 9 from 95 non - defective) is $C(5,1)\times C(95,9)=\frac{5!}{1!(5 - 1)!}\times\frac{95!}{9!(95 - 9)!}$.
The number of non - rejected samples is $C(95,10)+C(5,1)\times C(95,9)$.

Step3: Calculate the probability of rejection

The probability of rejection $P$ is $P = 1-\frac{C(95,10)+C(5,1)\times C(95,9)}{C(100,10)}$.
The probability that a lot contains exactly 3 defective items:
The number of ways to choose 3 defective items out of 5 is $C(5,3)=\frac{5!}{3!(5 - 3)!}=10$.
The number of ways to choose 7 non - defective items out of 95 is $C(95,7)=\frac{95!}{7!(95 - 7)!}$.
The number of samples of size 10 with exactly 3 defective items is $C(5,3)\times C(95,7)$.
The probability that a sample of size 10 has exactly 3 defective items is $\frac{C(5,3)\times C(95,7)}{C(100,10)}$.

Regarding the customer's decision, the sampling plan has some risks. There is a chance of rejecting a lot with a small number of defectives (Type I error) and accepting a lot with a relatively large number of defectives (Type II error). It is a trade - off between the cost of inspecting more items (which would reduce these errors) and the cost of accepting or rejecting lots.

Answer:

The probability of lot rejection is $1-\frac{C(95,10)+C(5,1)\times C(95,9)}{C(100,10)}$, the probability of exactly 3 defective items in the sample is $\frac{C(5,3)\times C(95,7)}{C(100,10)}$, and the customer's decision has associated Type I and Type II error risks with a trade - off in sampling costs.