# Combining $n$ probability observations of an object for higher probability

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;

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

While;

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

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 $$\dfrac{p_A(a_1)p_A(a_2)p_A(a_3)}{p_A(a_1)p_A(a_2)p_A(a_3)+(1 - p_{A}(a_{1}))(1 - p_{A}(a_{2}))(1 - p_{A}(a_{3}))}$$
• 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. – Xi'an Jan 14 '17 at 13:36