Bayes Theorem probability Past Exam Paper Question -
Prof. Smith is crossing the Pacific Ocean on a plane, on her way to a conference.
The Captain has just announced that an unusual engine fault has been
signalled by the plane’s computer; this indicates a fault that only occurs once in
10,000 flights. If the fault report is true, then there’s a 70% chance the plane
will have to crash-land in the Ocean, which means certain death for the passengers.
However, the sensors are not completely reliable: there’s a 2% chance of
a false positive; and there’s a 1% chance of the same fault occurring without the
computer flagging the error report.
Question 
Formulate this problem in terms of conditional probabilities of outcomes, existence of
a fault and whether or not it is reported and use Bayes’ rule to compute Prof. Smith’s chances of survival. 
My Attempt
P(Fault) - 0.0001
P(Crash | Fault) - 0.7
P(FalsePositive | Fault) - 0.02
P(NoReport | Fault) - 0.01
I have no idea what to do next, every example I look at seems  a lot easier than this. Could someone help me out? 

The first thing is to find what conditional probability they want.  This is usually the probability of what you want to know given what information you have.  In this problem we want to know whether you survive given that the computer said we have an engine fault $P(S \mid C = f)$ where f is fault and n is no fault.
Now we have to interpret each statement in the problem as either saying a conditional probability or an unconditional probability.  Because this is a conventional Bayes theorem problem there there is only going to be one unconditional probability.
probability of engine fault: $P(F) = .0001$.
probability of death given an engine fault: $P(D \mid F) = .7$
probability of computer giving false positive: $P(C = f \mid N) = .02$
probability of computer giving false negative: $P(C = n \mid F) = .01$
Now we use Bayes theorem
\begin{align*}
P(S \mid C = f) & = \frac{P(C=f \mid S) P(S)}{P(C = f)}
\end{align*}
We don't know the values on the right hand side so we need to consider cases (or use the partition theorem).  We are going to consider the two cases that there is actually a computer fault $F$ or there is not a computer fault $N$
\begin{align*}
P(S) & = P(S \mid F)P(F) + P (S \mid N)P(N) = .3*.001 + 1*.999 \\
P(C = f) & = P(C = f \mid F)P(F) + P(C=f \mid N)P(N) = (1-.01)*.001+.02*.999\\
P(C = f \mid S) & = P(C=f,S)/P(S) \\
& = P(C=f \mid F)P(S\mid F)P(F) /P(S) + P(C=f \mid N)P(S \mid N)P(N) /P(S) \\
& = (1-0.01)*.3*.001/.9993 + .02*1*.999/.9993
\end{align*}
In that last step I used that fact the events $C=f$ and $S$ are conditionally independent given $F$ or $N$.  Whether the computer flagged a fault and you survived are obviously not independent by themselves but if you know that a computer fault happened or not then they are independent.
Plugging the values in we get that $P(S \mid C = f) \approx .9986$.  This answer makes sense intuitively because even though false positives don't happen very much, computer faults happen even less in general and most people survive when flying in a plane.
