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)=

=0.3

Also P(B)=0.3.

Now the lot has n-1 components and 0.3n-1 of them are defective.

Thus, we have

P(B|A)=

=

=

=-

=0.3-

Now for n=10,

P(B|A)=0.3-

=0.3-0.07778

=0.2222

And for n=10,000,

P(B|A)=0.3-

=0.3-0.00007

=0.29993

Hence,P(B|A) at n=10 is 0.2222 and P(B|A) at n=10,000 is 0.29993.