Having trouble with this Probability Problem with a conditional

Hello everyone,

I'm taking statistics for the first time in my life and started probability and am trying to get the hang of these word problems, but I running into issues trying to apply everything I've learned so far (multiplication rule, addition rule, etc.). Anyway, here goes the problem.

At a glass manufacturing plant, crystal stemware is inspected for correct dimensions and shape. After a lengthy study, the factory is known to produce 15% defectives. Most of these pieces are reshaped or discarded. Suppose two pieces are randomly selected for inspection.

a. What is the probability that both pieces are defect free?

Okay, for a I tried to use the conditional rule and had: p(A inters B) / p(b) where p(a) = .85 that are not defective and p(b) are the 15% that are defective.

So, I calculated this out as: (.15 * .85)/.15 and got .85.............This could be the right answer or I just don't entirely know if my approach to things are right.

b. What is the probability that neither piece is defect free?

I did the conditional property again on this and switched things, so this time I had: (.15 * .85)/ .85 and got .15...........Again this is probably wrong, because it seems to simple.

c. Suppose at least one of the pieces has a flaw. What is the probability that BOTH are defective?

Lastly, I was not sure at all what to do here. I know the sample size is shrunk somehow I just can't think or understand how to do the problem wth the tools I've got. Plus, the examples given in the book were simpler than this. If anyone could point me in the right direction if inform me on how to do these, I would certainly appreciate it. I feel really stupid right now. I can do calculus but not these probability type word problems (Thinking)

Re: Having trouble with this Probability Problem with a conditional

Use the binomial distribution.

$\displaystyle P(X=x) = \binom{n}{x}p^x(1-p)^{n-x}=\dfrac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$

where n= total number of trials, p=probability of success which in this case is 0.85, x= number of successes

a. What is the probability that both pieces are defect free?

Use n=2, x=2, and p=0.85 in the above equation

b. What is the probability that neither piece is defect free?

Use n=2, x=0, and p=0.85 in the above equation

c. Suppose at least one of the pieces has a flaw. What is the probability that BOTH are defective?

find P(X=1)+P(X=2) using the same equation

Re: Having trouble with this Probability Problem with a conditional

Wow that is really different. In the book I've been using, we haven't learned anything like that formula yet. It looks like a great way to do things though. Is there another way to do this problem without the binom formula?