I've got $n$ observations of an object corresponding to $n$ readings;

$$a_{1}, a_{2}, \ldots, a_{n}$$

where $p_{A}(a_{j})$ is the probability that observed object in reading $a_{j}$ is 'A'.

Now, as all my readings are of the same thing, I would like to utilize all of them to find the probability that the thing I'm observing is indeed 'A'.

I am convinced that the method I am currently using is invalid, namely;

$$p_{A} = 1 - [(1 - p_{A}(a_{1})) \cdot (1 - p_{A}(a_{2})) \cdot \ldots \cdot (1 - p_{A}(a_{n}))]$$

By the example, that;

$$p_{A} = 1 - [(1 - 0.7) \cdot (1 - 0.8) \cdot (1 - 0.6)] = 0.976$$


$$p_{(1-A)} = 1 - [(1 - 0.3) \cdot (1 - 0.2) \cdot (1 - 0.4)] = 0.664$$

And thus that $p_{A} + p_{(1-A)} \neq 1$.

EDIT: New values and percentages.


The probability $$p_{A} = 1 - [(1 - 0.7) \cdot (1 - 0.8) \cdot (1 - 0.6)] = 0.976$$ corresponds to the probability of getting an A at least once during the three experiments, since $$(1 - 0.3) \cdot (1 - 0.2) \cdot (1 - 0.4)$$ is the probability to never get an A. This explains why your $P_A$ and $P_{1-A}$ do not sum up to one.

If you further assume that all readings take the same value, you need to specify a joint distribution on the experiments, because you only define the marginals through $p_A(a_i)$.

If instead you want to condition on the fact that the outcome is made of identical draws, all $A$'s or all $A^c$'s, the probability for getting all $A$'s then writes as $$\dfrac{p_A(a_1)p_A(a_2)p_A(a_3)}{p_A(a_1)p_A(a_2)p_A(a_3)+(1 - p_{A}(a_{1}))(1 - p_{A}(a_{2}))(1 - p_{A}(a_{3}))}$$

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  • $\begingroup$ I see that now, but then how would I go about combining my readings? - My intuition is that if multiple subsequent readings all point towards the same answer, with a high probability, then this strengthens the evidence for that answer being the correct one. $\endgroup$ – Skeen Jan 13 '17 at 19:07
  • $\begingroup$ I have absolutely no idea how to specify a joint distribution function. Help. I need help... silent crying $\endgroup$ – Skeen Jan 13 '17 at 22:21
  • $\begingroup$ @Skeen: What I mean is that the problem is incomplete as stated and that there is no answer unless you precise how the three draws are implemented. The specification of three marginal distributions is not compatible with the notion that the three draws are identical, i.e., all A's or all not A's. $\endgroup$ – Xi'an Jan 14 '17 at 9:47
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    $\begingroup$ Okay, I'll try to specify the problem. I'm monitoring the state of an object, and processing the raw data into a reading as described here (by using machine learning). The probabilities described are the output of the machine learner. The readings are independent. $\endgroup$ – Skeen Jan 14 '17 at 13:23
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    $\begingroup$ The formula describes the conditional probability of $A$ given the event "all $A$'s or all $A^c$'s". Your "specification" does not tell how the three experiments are related, hence does not allow for an answer to your question. I thus vote for the closing of the question. $\endgroup$ – Xi'an Jan 14 '17 at 13:36

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