Bayes Theorem probability

Question

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?

Answer 1

$P(F) = 0.0001$
$P(\lnot F) = 0.9999$
$P(+ve \; | \; \lnot F) = 0.02$
$P(-ve \; | \; F) = 0.01$

$P(+ve \; | \; F) = 1 - P(-ve \;|\; F) = 0.99$
$P(-ve \; |\; \lnot F) = 1 - P(+ve \;|\; \lnot F) = 0.98$

By Law of Total Probability (LOTP):
$P(+ve) = P(+ve \;|\; F)\;P(F) + P(+ve\;|\;\lnot\;F)\;P(\lnot\;F)$
$\quad\quad\quad\quad\quad\;\; = 0.99(0.0001) + 0.02(0.9999)$
$\quad\quad\quad\quad\quad\;\; = 0.020097$

By Bayes Theorem:
$P(F\;|\;+ve)=P(+ve\;|\;F)P(F)\;/\;P(+ve)$
$\quad\quad\quad\quad\quad\quad\quad\quad\;\; \approx 0.00492611$

From the Tree Diagram, Bayes and LOTP with extra conditioning: $P(Crash \;|\; +ve) = P(Crash \; | \; F,\; +ve) \; P(F \;|\; +ve) + P(Crash \;|\; \lnot F, +ve)\;P(\lnot F \;|\; +ve)$
$\quad\quad\quad\quad\quad\quad\quad\;\;=0.7\;(0.00492611)+0\;(0.9950738916)$
$\quad\quad\quad\quad\quad\quad\quad\;\;\approx 0.003448$

$P(Survive\;|\;+ve)=1-P(Crash\;|\;+ve) \approx 0.996552$

Answer 2

Let $R=$"A fault is reported", $F=$"A fault occurs" and $C=$"The plane must crash-land". Then a false positive is $R\cap F^c$, a non-reported fault is $R^c\cap F$. We are given $P(F)=1/10000$, $P(C|F\cap R)=70/100$, $P(R\cap F^c)=2/100$ and $P(R^c\cap F)=1/100$. We want to calculate $P(C^c)=1-P(C)$. By Baye's rule, $P(C)=\frac{P(C|F\cap R)P(F\cap R)}{P(F\cap R|C)}$.But $P(F \cap R)=P(F)P(F|R)=98/1000000$, and $P(F \cap R|C)=98/100$. Hence $P(C^c)=1-(70.98.100)/(100.1000000.98)=1-7/100000$

Answer 3

One mistake you made initially was that the .02 was not P(Report|Fault), but rather P(Report|NoFault), as that is essentially what a false positive is.

We want to find P(Crash|Report), which is equivalent to P(Crash|Fault) * P(Fault|Report), and we know P(Crash|Fault). This means that we need to find P(Fault|Report).

To do this, we use Baye's theorem, that states that:

P(Fault|Report) = P(Report|Fault) * P(Fault)/P(Report).

We can find each of these:

P(Report|Fault) = 1 - P(NoReport|Fault)

P(Fault) you already know

P(Report) = P(Report|NoFault) * P(NoFault) + P(Report|Fault) * P(Fault)

And you can just plug the values in from there.

Answer 4

When confused, it is useful to break the problem into parts, and work through a fictitious scenario.

The prof will survive all false alarms and $30\%$ of correct alarms

Suppose the prof takes $1,000,000$ trips (to avoid decimals)

A fault is likely to occur $100$ times, of which $99\% \;\; or\;\;99$ would sound the alarm,
and not occur $999,900$ times, of which $2\%\;\; or\;\; 19,998\;$ would sound false alarms

Thus of the $99+19,998 = 20097$ times the alarm sounds,
the prof will survive in $19,998+ 30\%$ of $99 = 20027.7$ times

and P(prof survives) $= \dfrac{20027.7}{20097} = 99.6551724..\%$

Added

Once you have got the scenario clearly mapped out, you should easily be able to put it in the usual mould for Bayes' rule. The important thing is to first get it clearly into your head.

PS:

To match with terminology in main answer,

P(Correct alarm) =P(fault)*P(alarm|fault) = $\dfrac1{10,000}\times(1-0.01) = \dfrac{99}{1,000,000}$

P(False alarm) = P(no fault)*P(alarm|no fault) $= \dfrac{9999}{10,000}\times(0.02) = \dfrac{19,998}{1,000,000}$

P(survive|alarm) $$= \dfrac{\text{P(survive|false alarm)*P(false alarm) + P(survive|correct alarm)*P(correct alarm)}}{\text{P(false alarm) + P(correct alarm)}}$$

Cancelling out the denominators, $$P(survive|alarm) =\dfrac{1*19,998+0.3*99}{19,998+99}= 99.6551724..\%$$

Answer 5

So we are given the information:

$P(\text{fault}) = \frac{1}{10000}$, $P(\text{crash}|\text{fault}) = 0.7$, $P(\text{yes}|\neg \text{fault}) = 0.02$,$P(\text{no}|\text{fault})=0.01$

where yes is the event that the signal goes off, and no is the event that the signal doesn't go off.

We know that:$$P(\text{crash}|\text{yes}) + P(\neg\text{crash}|\text{yes}) = 1$$ $$P(\text{crash}|\text{yes}) = P(\text{crash}|\text{fault}) P(\text{fault}|\text{yes})$$ $$P(\text{fault}|\text{yes}) = \frac{P(\text{yes}|\text{fault})P(\text{fault})}{P(\text{yes})}$$ $$P(\text{yes}) = P(\text{yes}|\text{fault})P(\text{fault}) + P(\text{yes}|\neg\text{fault})P(\neg\text{fault})$$

With these couple of equations, (the main one being the application of Bayes' theorem), we can work out $P(\neg\text{crash}|\text{yes})$, which turns out to be greater then $99$%.

Answer 6

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.

Answer 7

Bayes law: $P(A|B) = \dfrac{P(B|A) P(A)}{P(B)}$

$P(Fault|Positive\,Indiction) = \dfrac{P(Positive Indication | Fault) P(Fault)}{P(Positive\,Indication)}$

P(Positive Indication | Fault) = 0.99

P(Fault) = 0.0001

P(Positive Indication) is a little trickier. So, lets make this table.

$\begin{matrix} &Fault&No Fault&Total\\ Positive&0.99*0.0001&0.02*0.9999&0.020097\\ Negative&0.01*0.0001&0.98*0.9999&0.979902\\ Total &0.00001&0.9999\end{matrix}$

$P(Fault|Positive\,Indiction) = \frac{0.000099}{0.020097} = 0.0049$

P(Survival) = 1-P(Death) = $1 - 0.0049*0.70 = 0.9965$