Assume that you produce computer software.  After the software is
produced you have people test it.  The people running the test have an
accuracy rate of 95%.  This means that if there is a problem that they
discover it in 95% of the cases; 5% of the time they fail to discover
it. It also means that if the  software is okay 95% of the time they
realize this but 5% of the time they declare a piece of software that
is okay to be defective.
Here now the question: Assume 10,000 units of software are tested.  Of
those that the people declare to have a problem, what proportion
really do have a problem?  Based on past experience 1% of all software
has a problem.  Show your reasoning and explain your reasoning.

Hi again k9queen!
In order to answer this question, recall first the definition of
conditional probability. If A and B are events, then:

P(A/B)=P(A and B)/P(B).

Let'see the information we have from the problem. We know that, if the
software is defective, there is a 0.95 probability that the tester
will say it's defective. Call the events:

D: The software is defective
OK: The software is not defective
SD: The tester says the software is defective
SOK: The tester says the software is not defective

Then we know from the problem that:

P(SD/D) = 0.95 = P(SD and D)/P(D)

and

P(SOK/OK) = 0.95 = P(SOK and OK)/P(OK)

This is the information we have. The problem is to determine the
probability that a piece of software is defective given that the
tester said it's defective. In other words, we must find:

P(D/SD) = P(D and SD)/P(SD)

Thus we must find P(D and SD) and P(SD). Once we know those
probabilities, we get the answer by dividing them.

Obtaining P(D and SD) is easy. Recall that we know that

0.95 = P(SD and D)/P(D)

and we also know that P(D)=0.01. Therefore, P(SD and D) (which is
obviously equal to P(D and SD) ), is basically:

0.95 = P(SD and D)/0.01
0.0095 = P(SD and D)

Now we need to find P(SD). We can decompose P(SD) in the following
way:

P(SD) = P(SD/D)*P(D) + P(SD/OK)*P(OK)

As you can see, the first term of this sum is P(SD and D), while the
second term is P(SD and OK). When summing them, we get the marginal
probability P(SD), which is the number we need. From the previous
lines, we already know that the first term is 0.0095. What about
P(SD/OK)*P(OK)? Since P(SOK/OK)=0.95, then P(SD/OK)=0.05. Also, since
P(D)=0.01, then P(OK)=0.99. Therefore, P(SD/OK)*P(OK)=0.0495. So,

P(SD) = 0.0095 + 0.0495
P(SD) = 0.059

Finally,

P(D/SD) = P(D and SD)/P(SD) = 0.0095/0.059 = 0.16

So we found that a proportion 0.16 of those which are reported as
defective are actually defective.


Google search strategy
conditional probability
://www.google.com.ar/search?hl=es&ie=UTF-8&oe=UTF-8&q=conditional+probability&meta=


I hope this helps! If you have any doubts regarding my answer, please
request a clarification before rating it; otherwise I await your
clarification and final comments.

Best wishes!
elmarto