Where am I going wrong in computing the probability of detecting fraudulent machines?

Question

Say I have many voting machines and $5\%$ of them are fraudulent. What's the probability that I extract one fraudulent machine? That's just $5\%$. What's the probability that I extract two legit machines? That's the complementary event, so it's $95\%$ times $95\%$. The binomial distribution is being used here.

I think that in these answers I'm assuming that when I take a machine at random I'm able to tell with $100\%$ probability whether it is legit or it's fraudulent. I'd like now to assume now that I have a certain probability of telling the difference.

When a machine is legit, I should have $100\%$ probability of telling so. (Otherwise it's human error: getting a perfect machine by mistake and saying it's fraudulent.) But it may happen that a fraudulent machine is so well prepared that I will not find out. Let's say I detect fraudulent machines $70\%$ of the times.

What's the probability that I take a fraudulent machine and detect it as such? Now there are two events going on here. The first is extracting a fraudulent machine, which is $5\%$. The second is being able to tell the machine is indeed fraudulent, which is $70\%$.

Are these two events independent? Very strictly speaking, I think they are not. (If I'm really good at detecting frauds, criminals could get better at hiding their frauds.) But I will assume they're independent.

What's the probability that I take 3 machines at random and detect at least one of them as fraudulent? I think it's better to take the complementary event. What's the probability of taking 3 machines and they're all legit? $$ \begin{align} P(M{=}L \cap M{=}L \cap M{=}L) &= P(M{=}L) (P(M{=}L \cap M{=}L)\\ &= P(M{=}L) P(M{=}L) P(M{=}L)\\ &= (0.95)^3\\ \end{align} $$ So the desired probability is $1 - 0.95 = 0.05$. But this can't be right. I only detect fraudulent machines $70\%$ of the times. The probability here should be less than $0.05$. Can you lead me in the right direction here?

Answer 1

First problem: your $70\%$ is really the probability that you detect a machine as fraudulent given that it is fraudulent. It is a conditional probability. It makes no sense to ask whether this event is independent from the machine being fraudulent or not (the question "is $A\mid B$ independent from $B$" makes no sense). But this has no impact on your computation.

Second problem: in the final computation you wrote $1-0.95$ instead of $1-0.95^3$.

Third problem, most important: the complementary event to "at least one machine detected as fraudulent" is no "all legit" but "all detected as legit". I think that given this remark you will be able to correct the computation by yourself.

Answer 2

In order for a machine to be detected as fraudulent we need two things to happen: it has to actually be fraudulent AND you must detect it as such. Hence the probability that a random machine is detected as fraudulent is $.05\times .7=.035$. The probability that none of three are detected as fraudulent is then $$\left( 1-.035 \right)^3\approx 0.899$$

Thus the answer you seek is $\boxed {0.101}$.

Note: I don't know why you think the answer should be less than $.05$ If you detect fraud flawlessly the answer would be $1-.95^3\approx 0.143$ .

For intuition: if you neglect the possibility of multiple fraudulent machines then the probability of seeing one fraud out of three is $3$ times the single machine probability. Hence about $.15$ if you detect perfectly and $.105$ if you detect at $70\%$. Note that both approximations are pretty good. That's because the probability of there being multiple frauds in your (small) sample is very low. If you were looking at more machines this approximation would become unusable.