I'm getting about every other question right, and since the back of the book only has answers, not whole solutions, I can't tell what I'm doing wrong. Here is the question:

In the process of producing engine valves, the valves are subjected to a first grind. Valves whose thicknessses are within the specifications are ready for installation. Those valves whose thicknesses are above the specifications are reground, while those whose thicknesses are below the specification are scrapped. Assume that after the first grind, 70% of the valves meet the specifications, 20% are reground, and 10% are scrapped. Furthermore, assume that of those valves that are reground, 90% meet the specifications, and 10% are scrapped.

a) Find the probability that a valve is only ground once.

P(valve is in spec) = P(A) = .7
P(valve is under spec) = P(C) = .1

P(A U C) = P(A)+P(C) = .8

Which is the right answer. Easy enough.


*b) Given that a value is ground only once, what is the probability that it is scrapped?

I was a bit confused on thise one.
The definition of conditional probability is
P(A|B) = P(A ^ B)/P(B)

So, P(Under Spec | Ground Once) = P(Under Spec ^ Ground Once)/P(Under Spec)
=(.7 * .1)/.1 = .7,

but the correct answer seem to be

(.7 + .1)/.1 = .125

which intuitively makes sense to me, because we're taking the probabiliy of it being under spec out of a small sample space.

I'm not sure if I'm using the P(A|B) formula incorrectly, or what.


c)Find the probability that the valve is scrapped

P(Under spec) = P(C) = .1
P(Over spec) = P(B) = .2
P(under spec after reground) P(E) = .1

P(Under Spec U (Over spec ^ reground under spec)) = P(C U (B ^ E))

=P(C) + P(B ^ E)
=P(C) + P(B)*P(E)
=.1 + .2*.1 = .12


*d)Given that a valve is scrapped, what is the probability that it was ground twice?

P(Scrapped) = P(S) = .12
P(Ground Twice) = P(Over spec) = P(B) = .2
(if it's over spec, it's going to be ground twice, whether or not the second grind is successful
P(S|B) = P(S^B)/P(B)
= P(S)*P(B)/P(B) , which doesn't seem correct, since that is just P(S).


There are a couple more parts to the question, but I'll wait to see if the feedback helps me solve them. Thanks for your time.

for part b consider what would happen to a group of 100 valves:

70 would pass inspection and 10 would be scrapped, the rest would be ground twice.

now to answer P(scrap|ground once) you need only count the number of scrap out of those ground once so it should be 10/80 = .125


now to your mistake:

"So, P(Under Spec | Ground Once) = P(Under Spec ^ Ground Once)/P(Under Spec) =(.7 * .1)/.1 = .7"

if your ^ means "and"

this should be: P(Under Spec | Ground Once) = P(Under Spec ^ Ground Once)/P(Ground once) and P(under^ground once) = .1, while P(ground once) = .8

cheers
jerry