Probability Question - Conditional Probability Quantity

Question

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.

Thanks in Advance.

Answer 1

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.

Answer 2

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

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

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