Answer to Question #122158 in Statistics and Probability for murtaza

Question

In each lot 200 items, two items are tested and the lot is rejected if either of the tested items is found
defective.
(a) Find the probability that a lot with k defective items is accepted.
(b) Suppose that when the production process malfunctions, 50 out of 100 items are defective.
In order to identify when the process is malfunctioning, how many items should be tested
so that the probability that one or more items are found defective is at least 99%?

Answer 1

(a) In each lot 200 items, there are k defective items and 200-k non-defective items. Then the probability that a lot with k defective items is accepted is

$$P(accepted)=\dfrac{\dbinom{k}{0}\dbinom{200-k}{2}}{\dbinom{200}{2}}=$$

$$=\dfrac{1\cdot\dfrac{(200-k)!}{2!(200-k-2)!}}{\dfrac{200!}{2!(200-2)!}}=\dfrac{(200-k)(200-k-1)}{200(199)}$$

(b) Let $X=$ the number of defective item: $X\sim Bin (n, p)$

Given $p=\dfrac{50}{100}=0.5$

The probability that one or more items are found defective is

$$P(X\geq 1)=1-P(X<1)=1-P(X=0)=$$ $$=1-\dbinom{n}{0}(0.5)^0(1-0.5)^{n-0}=1-(0.5)^n$$

$$P(X\geq1)\geq0.99=>1-(0.5)^n\geq0.99$$

$$2^n\geq100$$

$$2^6=64<100, 2^7=128>100$$

At least 7 items should be tested so that the probability that one or more items are found defective is at least 99%.