How to solve this task about Conditional probability and is my solution correct?

Question

I have the following task from a book.

In a factory components are build. Each components gets testest by a test. The sensitivity of this test is 90%, means out of all components that are broken the test identifies 90% correctly as broken. Out of all not broken components the test identifies 98% correctly as not broken. 2% of all components are broken (before they get testet.

a): If the test identifies a component as not broken, how high is the probability that this component is actually not broken?

b): If the factory produces a batch of 10000 components that all have been testet not broken, how high is the probability that at least one of those components will be actually broken even thought all tests were negative?

My Idee: a) $P(a) = 0,98 \cdot 0,98 / (0,98 \cdot 0,98+0,02*0,1) = 0.9979218$

b) P(b) = 1- bin(n = 100,k =0, p = 0.002) = 0.181433 ^, with $p = 0.02 \cdot 0.10 = 0.002$

Is this correct?

Answer 1

This is how I would do a problem like this: Imagine 10,000 such components.

"2% of all components are broken".
So 0.02(10000)= 200 are broken the other 9,800 are not.

"out of all components that are broken the test identifies 90% correctly as broken". So 0.9(200)= 180 of the broken components are correctly identified as broken the other 20 are incorrectly identified as not broken.

"Out of all not broken components the test identifies 98% correctly as not broken." So 0.98(9800)= 9604 of the unbroken components are correctly identified as unbroken, the other 196 unbroken components are incorrectly identified as broken.

a) So a total of 180+ 196= 376 components are identified as broken of which 180 really are. If a component is identified as broken, the probability it really is broken is 180/376= 45/94.

b) A total of 20+ 9604= 9624 components are identified as not broken of which 9604 really are not broken. The probability that a component that was identified as unbroken really is unbroken. 9604/9624= 4802/4812= 2401/2406.

If 10000 components are identified as unbroken then (2401/2406)10000= 9979 actually are unbroken.

Answer 2

a) Your idea is almost right. The denominator, i.e. the given probability that the test indicates "not broken", is correct. Only the numerator is unfortunately wrong, you want the probability that the test indicates "not broken" although the component is "broken" and not "not broken" if it is actually not broken. The numerator is, therefore, $0.02*0.9$ and not $0.98*0.98$. Summarizing the probability is given by \begin{align*}P(a)=\frac{0,02*0,1}{0,98*0,98+0,02*0,1}\approx 0,0021.\end{align*} b) I would suggest solving this task as follows. Compute the counter-probability (all components are not broken given all the tests are negative) \begin{align*} (1-P(a))^{10000}\approx 1*10^{-9}. \end{align*} Now the probability that at least one component is broken is given by \begin{align*} 1-(1-P(a))^{10000}\approx99,9999999\%. \end{align*} You can also do it with Binomial-distribution, but the number of components $n$ should be 10000 by the description and the probability $p$ is the one from task a) since you already know that all tests were negative.