Probability of not detecting fire if two of the sensors don't work well

Question

I am trying to solve the following problem of probability:

"In a certain factory, there is an alarm system composed of 5 sensors that work independently to detect fire. The probability that a sensor works well, that is, warns people when there's fire is $0.8$. If two of the sensors don't work well, which is the probabilty of not detecting the fire?".

I have thought of Bayes $p(A|B)$ where A="Not detecting fire" and B="2 sensors don't work well". In this way, I have computed the probability as:

$$\frac{1*0.2^5}{\binom{5}{2}0.2^2*0.8^3}$$. However, the result of this isn't the one that appears in the solutions of the textbook because the solution should be $0.008$. Any help to approach the problem? Thanks.

Answer 1

We will assume that "sensor works" is synonymous to "detects fire". The question then becomes "given that at least two sensors don't work, what is the probability that none of the sensors work?".

The probability that two of the sensors don't work can be found via complementary counting by subtracting from unity the probability that all the sensors work and the probability that only one of the sensors doesn't work: $1 - (0.8)^5 - 5 (0.2) (0.8)^4$. The probability that none of the sensors work (leaving the fire undetected) is $(0.2)^5$. The desired conditional probability is

$$\frac{(0.2)^5}{1 - (0.8)^5 - 5 (0.2) (0.8)^4} \approx \boxed{0.001218}$$

EDIT: Clarification on why $0.008$ is a bad answer.

As @John Douma mentioned in the comments, the textbook answer was probably found by taking $(0.2)^3$. However, as @Stinking Bishop has pointed out, this assumes that we know that two specific alarms don't work. That is not the same question as what was posed, which is that given that some pair of alarms don't work, what is the probability that the fire is not detected.

Answer 2

Well, as John Douma pointed out in the answer, the probability is $(1-p)^3 = 0.008$. This is because the probability of a sensor failing is $1-p = 1-0.8 = 0.2$. Since you have 3 sensors (the other two are already faulty), we get $0.2^3 = 0.008$.
To get the same answer using $P(A|B)$, you need to use the formula properly $$P(A|B) = {P(A \cap B)\over P(B)} = {n(A \cap B)\over n(B)}$$ $n(A \cap B)$ is just the number of ways in which $2$ are permanently faulty and the remaining fail. That is $\binom{5}{2}(0.2)^5$. $n(B)$ is the number of ways you can choose which ones fail, so that is $\binom 52(0.2)^2$. Now apply the formula: $$P(A|B) = {\binom{5}{2}(0.2)^5\over \binom 52(0.2)^2} = 0.008$$ You've made two errors: You've considered the probability of all $5$ faulting in the numerator, but you are already given $2$ of them are faulty, so you should just choose the faulty ones (because you need to differentiate between which ones were already faulty and which faulted during the fire, otherwise you're undercounting). In the denominator, $B$ is just $2$ don't work well (so again, just choose them and make them faulty), you are not required to make the others error-free. (If you make the others error-free, $A$ and $B$ become disjoint).