Combining probability from multiple sensors Sorry for asking a relatively simple question, but I am new to this field.
There are two sensors, the first sensor $S1$ with an accuracy of $M\%$ and another $S2$ with an accuracy of $N\%$. The sensors are detecting an event $E$ (i.e., producing yes/no signals) with has a $Q$ probability of occuring.
If both the sensors report that the event has occured, i.e., generate a 'yes' signal at a given instance, what is the actual probability of the event having occured?
I guess the probability of getting a 'yes' signal from the sensors are obtained using Bayes' theorem, such that:
\begin{equation}
P(E|S1) = \frac{P(S1|E) \cdot P(E)}{P(S1)} = \frac{P(S1|E) \cdot Q}{\frac{M}{100}}
\end{equation}
\begin{equation}
P(E|S2) = \frac{P(S2|E) \cdot P(E)}{P(S2)} = \frac{P(S2|E) \cdot Q}{\frac{N}{100}}
\end{equation}
I am wondering if I can interpret $P(S1|E) = P(S1)\cdot P(E)$ and so on. I am unsure how to proceed with this.
Any help would be appreciated. Thanks.
 A: Here's one method.
We draw a causal diagram of the earthquake and the sensors. The model we assume is $S_1\leftarrow E \rightarrow S_2$. We see that the earthquake "causes" both the sensors, and we have independence of $S_1,S_2$ given the earthquake $E$.
Notation: we can assume $M,N$ as probabilities, instead of dividing by 100. Accuracy of $S_1$ is $M$. We will take this to mean $S_1$ goes off with probability $M$ when there is an earthquake and goes off with probability $1-M$ when there is no earthquake. (Similarly with $S_2$)

\begin{align*}
\mathbb P [E|S_1,S_2] &= \frac{\mathbb P [S_1,S_2|E]\mathbb P [E]}{\mathbb P [S_1,S_2]}\\
&= \frac{\mathbb P [S_1,S_2|E]\mathbb P [E]}{\mathbb P [S_1,S_2|E]\mathbb P [E]+\mathbb P [S_1,S_2|\overline E]\mathbb P [\overline E]}\\
\end{align*}
Now we use conditional independence of sensor 1 and 2.
\begin{align*}
\mathbb P [E|S_1,S_2] &= \frac{\mathbb P [S_1|E]\mathbb P [S_2|E]\mathbb P [E]}{\mathbb P [S_1|E]\mathbb P [S_2|E]\mathbb P [E]+\mathbb P [S_1|\overline E]\mathbb P [S_2|\overline E]\mathbb P [\overline E]}\\
\end{align*}
Now we substitute the values
\begin{align*}
\mathbb P [E|S_1,S_2] &= \frac{MNQ}{MNQ+(1-M)(1-N)(1-Q)}\\
\end{align*}
Example 1:
Taking some values for illustration: say M = 0.9, N = 0.9 and Q = 0.01, then
$\mathbb P [E|S_1,S_2] = \frac{0.9\times0.9\times0.01}{0.9\times0.9\times0.01+0.1\times0.1\times0.99} = 0.45$

Coming to your calculations for comparison
\begin{align*}
\mathbb P [E|S_1] &= \frac{\mathbb P [S_1|E]\mathbb P [E]}{\mathbb P [S_1|E]\mathbb P [E]+\mathbb P [S_1|\overline E]\mathbb P [\overline E]}\\
&=\frac{MQ}{MQ+(1-M)(1-Q)}\tag 1
\end{align*}
Similarly,
\begin{align*}
\mathbb P [E|S_2] &= \frac{NQ}{NQ+(1-N)(1-Q)}\tag 2
\end{align*}
For the given values in example 1, $\mathbb P [E|S_1]=\mathbb P [E|S_2] = 0.083$, so we have an improvement in accuracy given 2 alarms.
Also see here for a related question.
