Probability that the fish that set off metal detector is the one true fish

Question

I have exams in Machine Learning coming up and I need help answering this question.

There are a million identical fish in a lake, one of which has swallowed the One True Ring. You must get it back! After months of effort, you catch another random fish and pass your metal detector over it, and the detector beeps! It is the best metal detector money can buy, and has a very low error rate: it fails to beep when near the ring only one in a billion times, and it beeps incorrectly only one in ten thousand times. What is the probability that, at long last, you’ve found your precious ring?

This is my answer I worked out using Bayes rule:

Is this the right way to work out this type of question and is that somewhat the correct answer?

Answer 1

I think you have mixed up a few of your conditional probabilities. If the detector fails to beep over the ring only one in a billion times, then it does beep the rest of the billion times. Similarly, if it beeps over he wrong fish one in ten thousand times, then it doesn't beep on the rest of those ten thousand wrong fish. So you should have \begin{align} P(B|\tilde{A}) &= \frac{1}{10000}\\ P(\tilde{B}|\tilde{A}) &= \frac{9999}{10000}\\ P(\tilde{B}|A) &= \frac{1}{1000000000}\\ P(B|A) &= \frac{999999999}{1000000000} \end{align} which is different from what you do have. Using these values gives \begin{align} P(B) & =\frac{999999999}{1000000000}\frac{1}{1000000} +\frac{1}{10000}\frac{999999}{1000000} \\ & =\frac{999999999}{1,\!000,\!000,\!000,\!000,\!000} +\frac{999999}{10,\!000,\!000,\!000} \\ & = \frac{100,\!999,\!899,\!999}{1,\!000,\!000,\!000,\!000,\!000} \end{align} and \begin{align} P(A|B) & = \frac{\frac{999999999}{1000000000}\frac{1}{1000000}} {\frac{100999899999}{1000000000000000}} \\ & = \frac{999,\!999,\!999}{1,\!000,\!000,\!000,\!000,\!000} \frac{1,\!000,\!000,\!000,\!000,\!000}{100,\!999,\!899,\!999} \\ & = \frac{999,\!999,\!999}{100,\!999,\!899,\!999}\approx 0.0099 \end{align} Or slightly less than $1$%. To check, we can make a quick estimate of the answer. There are almost one million fish that don't have the ring. The detector gives a false positive one in ten thousand times. So, if you were to test every fish in the lake, you would get approximately $100$ beeps. Of all the fish that trigger beeps, only one has the ring, which means that the detector would only be right about once in every hundred beeps, or $1$% of the time.

In regards to the other part of your question, yes, your methods were correct. It was just your starting values that were wrong.

Answer 2

Surely the probability of the detector beeping if you have found the fish is 999999999/1000000000 not 9999/10000 Whereas the probability of the detector not beeping when the fish is not found is 9999/10000 not 999999999/1000000000?

Answer 3

You have $2$ cases in which the detector will beep:

$1:$ You have found the fish with the ring and the detector beeps. Probability:

$$\frac{1}{10^6} \cdot (1-\frac{1}{10^9}) = 9.99999999 \cdot 10^{-7}$$

$2:$ You have not found the fish with the ring but the detector beeps anyway. Probability:

$$\frac{10^6 -1}{10^6} \cdot \frac{1}{10^4} = 0.0000999999$$

To get the probability that you catched the correct fish under the condition that your detector beeped is then

$$\frac{\frac{1}{10^6} \cdot (1-\frac{1}{10^9})}{\frac{1}{10^6} \cdot (1-\frac{1}{10^9})+\frac{10^6 -1}{10^6} \cdot \frac{1}{10^4}}= \frac{1001001}{101101001} \approx 0.9901 \%,$$ so slightly less than $1\%$.