In a lot of n components, 30% are defective. Two components are drawn at random and tested. Let A be the event that the first component drawn is defective, and let B be the event that the second component drawn is defective. For which lot size n will A and B be more nearly independent: n = 10 or n = 10,000? Explain.
Step 1 of 3:
Here it is given that 30% of the components are defective in a lot of n components. Two components from the lot are selected randomly and are tested.
A denotes the event that the first component drawn is defective and B denotes the event that second component drawn is defective.
We have to check for which value of n the two events A and B are independent, either for the lot sizes n=10 or for n=10,000.
Step 2 of 3:
It is given that 30% of the components are defective. Thus P(A)=
Now the lot has n-1 components and 0.3n-1 of them are defective.
Thus, we have
Now for n=10,
And for n=10,000,
Hence,P(B|A) at n=10 is 0.2222 and P(B|A) at n=10,000 is 0.29993.