You are programming a demining robot. As the robot drives along, the prior probability of a mine being in its immediate vicinity is 0.001 The robot is equipped with a mine detecting sensor which sounds a bell if it detects a mine nearby. If there is a mine nearby, then the probability of the sensor ringing the bell is 0.95. If there is no mine nearby, the probability of the sensor ringing the bell is 0.01
You now fit a new sensor to the robot. This uses a different method to detect mines from that used by the first sensor, and if it does detect a mine, it sounds a buzzer. If there is a mine nearby, then the probability of the new sensor sounding the buzzer is 0.90, and if there is no mine nearby, the probability of the sensor ringing the bell is 0.02. How would you use the ringing alarms from both sensors to compute the probability of there being a mine nearby?
With b = bell, m = mine, z = buzzer I have:
P(b,z|m)p(m) / P(b,z)
My only thought is that b and z are independent which would give:
P(b|m)P(z|m)P(m) / P(b)P(z)
P(b) = P(b|m)P(m) + P(b|¬m)P(¬m)
and likewise for z. I plug in values and get > 1, I'm not sure how to approach it
I am assuming that there is a typo above, and that on the new sensor, there is a probability of $0.02$ of the buzzer sounding, if there is no mine nearby.
I am also assuming that false positives or false negatives with respect to the bell are totally independent of false positives or false negatives with respect to the buzzer. That is, I am assuming (for example) that there is no atmospheric condition that would simultaneously cause both sensors to malfunction.
As I see it, the assumption in the above paragraph is justified, because otherwise, I don't see how the problem can be attacked.
I am unsure about the informality of the following approach, but the approach makes sense to me.
Let $E_1$ denote the event that there is a mine in the vicinity.
Let $E_2$ denote the event that the bell rings.
Let $E_3$ denote the event that the buzzer sounds.
There are $8$ possibilities, re either event $E_k$ occurs or it doesn't
for $k \in \{1,2,3\}$.
What you do is assign a probability to each of the $8$ possibilities occurring.
Then, you consider the $4$ known alternatives re either event $E_2$ occurs or it doesn't and either event $E_3$ occurs or it doesn't.
Based on your computation of the $8$ probabilities of how the events may unfold, you use Bayes Theorem to assign a probability to event $E_1$ occurring, based on each of the $4$ alternatives with respect to events $E_2$ and $E_3$.
$\underline{\text{Case 1: the bell rings and the buzzer sounds.}}$
Probability of mine is $0.001$.
If there is a mine, probability of bell is $0.95$.
If there is a mine, probability of buzzer is 0.90.
Therefore, probability of mine and bell and buzzer is
$P_1 = 0.001 \times 0.95 \times 0.90.$
Probability of no mine is $0.999$.
If there is no mine, probability of bell is $0.01.$
If there is no mine, probability of buzzer is $0.02.$
Therefore, probability of no mine, and bell and buzzer is
$P_2 = 0.999 \times 0.01 \times 0.02.$
Therefore, in Case 1, which is focusing exclusively on the situation where both the bell and buzzer sound, the chance of a mine is
$$\frac{P_1}{P_1 + P_2}.\tag 1$$
The other $3$ alternatives, re the bell and the buzzer are analyzed in the same way as that done in Case 1. You end up with a computation formatted the same way as that of (1) above.