Conditional Probability: an incorrect error detection

[Catchfire]

Homework Statement

From Mathematical Statistics and Data Analysis 3ed, Rice

1.8 #61
Suppose chips are tested and the probability they are detected if defective is 0.95, and the probability they are declared sound if they are sound is 0.97. If 0.005 of the chips are faulty. What is the probability that a chip that is declared faulty is sound?

Homework Equations

P(A|B) = P(A\capB) / P(B)

P(A) = \SigmaP(A|Bi)P(Bi)

The Attempt at a Solution

Let D- be the event a fault is detected
Let D+ be the event no fault is detected
Let Df be the event a chip is faulty
Let S be the event a chip is sound

P(D-|Df) = 0.95
P(D+|S) = 0.97
P(Df) = 0.005
P(S) = 1 - P(Df) = 0.995

Find P(S|D-) (the answer given is 0.86)

P(S|D-) = P(D-\capS) / P(D-) = P(D-|S)P(S) / P(D-)

So here's where I've been stuck.
First I'd like to know if I've translated the problem correctly.
Secondly how do I find P(D-|S) and P(D-) or am I going about this the wrong way?

 

[D H]

P(D-|S) is simple. Suppose a sound chip is tested. There are two possible outcomes of this test, that a defect is or is not detected. These events are D+|S and D-|S. Since you know the probability of one of those outcomes, obtaining the other is trivial.

P(D-) is also simple. Use the formula of total probability, the second of your relevant equations.