bayesian prob question? Question:

Imagine you work at a computer repair shop. Over the years you have
become experienced at diagnosing problems associated with different
types of hardware. In particular, when a “TX1000” brand computer stops
working, it is due to a cooling fan problem $30\%$ of the time. If the
cooling fan is broken, there is a $45\%$ probability that the hard drive
is also broken. If the cooling fan is okay, there is only a $5\%$ chance
that the hard drive is broken.
A customer brings you a computer that is not working. After a quick
investigation, you determine that the hard drive is broken. What is
the probability that the computer has a bad cooling fan?

I'm not sure if I am solving this question right. I am assuming its a bayesian probability question. Assuming that I have calculated these numbers:
$.45\cdot.3= 0.135\tag{1}$
cooling fan and hard drive broken
$.55\cdot.3 =0.165\tag{2}$
cooling fan broken, hard drive ok
$$.05\tag{3}$$
cooling fan ok, hd broken
$$.95\tag{4}$$
if cooling fan not ok, hard drive broken
So far I've calculated:
$$A \cap B= \frac{.135\cdot.05}{(.135\cdot.05)+(.165\cdot.95)}=0.0412844\cdot100\%=4.13\%$$ chance of a bad cooling fan.
Something seems conceptually wrong about this though. I feel like I'm messing up the calculation.
 A: I think you went a little off track in your calculation. I'll start at the beginning and work through so you can see where the difference occurs:
First, some definitions of events: H = Hard drive broken, C=cooling fan broken
Second: What do we know:
$P(C)=.3, \\P(C^c)=1-P(C)=.7,\\P(H|C)=.45,\\P(H|C^c)=.05$
Third: What do we want to know? $P(C|H)$
You are correct that this is a Bayes theorem question. We can assemble the pieces below using the information we have:
$P(C|H)=\frac{P(H\cap C)}{P(H)}=\frac{P(C)P(H|C)}{P(C)P(H|C)+P(C^c)P(H|C^c)}=\frac{(.3)(.45)}{(.3)(.45)+(.7)(.05)}=\frac{.135}{.17}\approx 79\%$
You went off track in the denominator, where you are multpilying joint probabilities instead of an unconditional times a conditional to get the "partitions" of event H. Notice that in my denominator we end up with $P(H\cap C)+P(H\cap C^c)$ since these two events are mutually exclusive and exhaustive (the fan is either broken or it isn't), you can add them to get the total probability of H (i.e., law of total probability)
This makes intuitive sense as well, as it is far more likely that the hard drive would be broken if the fan were broken (Likelihood ratio/bayes factor of $\frac{.45}{.05} = 9$)  
