# Probability Question - Conditional Probability Quantity

A particular fault occurs in a certain type of mechanical devices with a frequency of 8 in 1,000.

A screening test for this fault is developed such that

(i) if the fault is present, it is detected with a probability of 98% and

(ii) if the fault is absent, the test has a 5% probability of showing (incorrectly) that there is a fault.

(a) What is the chance that a device which is shown by the test to be faulty, is in fact faulty?

(b) What is the chance that a device which is shown by the test not to be faulty, is in fact faulty? In a batch of 1,000,000 devices, how many faulty devices can be expected to pass through this screening test?

I am unsure on how to answer the batch question of the last part, on how many faulty devices can be expected to pass through.

• Welcome to Cross Validated! If this is a self-study question (as it looks to be), we ask that you tag the question as 'self-study' and follow the suggestions here: stats.stackexchange.com/tags/self-study/info – Matt Krause Apr 11 '16 at 0:35
• Please note that it's not enough to add the tag, you need to also add your own thoughts on the problem. For example. if only the batch question of (b) is unclear to you, you could explain your solutions to (a) and the first question of (b) and then explain your initial thoughts on the batch question of (b). – Juho Kokkala Apr 11 '16 at 10:21

I believe that the second question is actually the more straightforward.

Out of 1 million devices, 8 one thousands will be faulty. 1,000,000 * (8 / 1,000) = 8,000 total faulty units.

Only 2% of these will not be detected (100% - 98%), so 8,000 * 0.02 = 160 faulty units passed the screening.

This is a classical problem of conditional probabilities.

Let's name the events...

$F$: device is faulty
$\overset{-}{F}$: device is not faulty
$D$: device is shown by the test to be faulty
$\overset{-}{D}$: device is shown by the test not to be faulty

notice that events $F$, $\overset{-}{F}$ and $D$, $\overset{-}{D}$ are complementary. Therefore...
$P(F)+P(\overset{-}{F})=1$
$P(D)+P(\overset{-}{D})=1$

What we know?

$P(F)=\frac{8}{1000}$

$P(D|F)=\frac{98}{100}$

$P(D|\overset{-}{F})=\frac{5}{100}$

Also...

$P(\overset{-}{F})=1-P(F)=1-\frac{8}{1000}=\frac{992}{1000}$

To solve the problem will need to calculate the probabilities of events $D$ and $\overset{-}{D}$.

$P(D)=P(D∩F)+P(D∩\overset{-}{F})=P(F)\times P(D|F)+P(\overset{-}{F})\times P(D|\overset{-}{F})$

$=\frac{8}{1000}\times \frac{98}{100}+\frac{992}{1000}\times \frac{5}{100}=\frac{5744}{10^5}$

$P(\overset{-}{D})=1-P(D)=1-\frac{5744}{10^5}=\frac{94256}{10^5}$

The questions...

(a)
The chance that a device which is shown by the test to be faulty, is in fact faulty

$=P(F|D)=\frac{P(F∩D)}{P(D)}=\frac{P(D|F)\times P(F)}{P(D)}=\frac{\frac{98}{100}\times \frac{8}{1000}}{\frac{5744}{10^5}}=\frac{98\times8}{5744}=\frac{784}{5744}\approx13.65\%$

(b)
the chance that a device which is shown by the test not to be faulty, is in fact faulty

$=P(F|\overset{-}{D})=\frac{P(F∩\overset{-}{D})}{P(\overset{-}{D})}=\frac{P(\overset{-}{D}|F)\times P(F)}{P(\overset{-}{D})}=\frac{(1-P(D|F))\times P(F)}{P(\overset{-}{D})}\\ =\frac{(1-\frac{98}{100})\times \frac{8}{1000}}{\frac{94256}{10^5}}=\frac{\frac{2}{100}\times \frac{8}{1000}}{\frac{94256}{10^5}}=\frac{16}{94256}\approx0.017\%$

In a batch of 1,000,000 devices, how many faulty devices can be expected to pass through this screening test?
This is the probability a device is faulty and the device passes the test multiplied by 1,000,000

$P(F∩\overset{-}{D})\times10^6=P(\overset{-}{D}|F)\times P(F)\times10^6=(1-P(D|F))\times P(F)\times10^6$

$=(1-\frac{98}{100})\times\frac{8}{1000}\times10^6=\frac{2}{100}\times\frac{8}{1000}\times10^6=16\times10=160$