P(tomorrow is the end of the world) =? 
Here we have $3$ prophets alpha, beta, gamma, they all predict that tomorrow is the end of the world. It's known that the accuracy of alpha's and beta's prediction is $90\%$, while that of gamma's is $4\%$. What is the probability that tomorrow is the end of the world?

Here's my calculations.
Let $A, B, C$ be alpha's, beta's and gamma's predictions.
P(tomorrow is the end of the world)
$= P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC)$
$= 0.9 + 0.9 + 0.04 - 0.81 - 0.036 - 0.036 + 0.0324$
$= 0.9904$
But this seems a bit too high, is my calculation correct? Thanks.
 A: I get 27/35, assuming independence. The two possibilities are that all three are right (.9 * .9 * .04) or all three are wrong (.1 * .1 * .96). The former divided by the sum gives 27/35 or about 77.1%.
If you don't assume independence -- for example, if beta just says whatever alpha says -- you can't reduce the problem to a single number without more information.
A: To me, the most sensible way to look at the probability that a prophet's prediction is true, is to look at the ratio of times when their predictions come true over the times when their predictions don't come true:
$$
\Lambda(A)=\frac{P(A\mid X)}{P(A\mid\neg X)}
$$
Extending the ideas used in this answer, and assuming that the prophets are independent, we get that
$$
O(W\mid A\cap B\cap C)=\Lambda(A)\,\Lambda(B)\,\Lambda(C)\,O(W)
$$
Where $O(X)$ is the odds of $X$ defined as
$$
O(X)=\frac{P(X)}{1-P(X)}
$$
We are given that
$$
\begin{align}
\Lambda(A)=\frac{.9}{.1}&=9\\
\Lambda(B)=\frac{.9}{.1}&=9\\
\Lambda(C)=\frac{.04}{.96}&=\frac1{24}\\
\end{align}
$$
Thus, adding the information from the prophets would increase the odds by a factor of $\frac{81}{24}$:
$$
O(W\mid A\cap B\cap C)=\frac{27}{8}O(W)
$$

This matches Robert Israel's answer if we convert from odds to probabilities:
$$
\begin{align}
P(W\mid A\cap B\cap C)
&=\frac{\frac{27}{8}\frac{P(W)}{1-P(W)}}{\frac{27}{8}\frac{P(W)}{1-P(W)}+1}\\
&=\frac{27P(W)}{27P(W)+8(1-P(W))}\\
&=\frac{.0324P(W)}{.0324P(W)+.0096(1-P(W))}
\end{align}
$$
A: The calculation is completely bogus.  The product formula is only true for independent events.  But here the three events are far from independent: either it is the end of the world or it isn't, so either they are all right or they are all wrong.
Let's try to put in some of the assumptions that are needed, and make a correct calculation with a Bayesian approach.  Suppose the three prophets are asked to predict whether a certain event $E$ will take place, and that event actually has probability $p$.
Suppose that whenever prophet $\alpha$ is asked to make a prediction of this
sort, his prediction is correct with probability $0.9$ (i.e. if the event actually happens, then with probability $0.9$ he predicted it would happen, and 
if it doesn't happen, then with probability $0.9$ he predicted it wouldn't).
Similarly for $\beta$ and $\gamma$, with probabilities $0.9$ and $0.04$ respectively.  Finally, suppose that the four events E, A = ($\alpha$ correct), B = ($\beta$ correct), C = ($\gamma$ correct), are independent.
The probability that the event will happen and all three say it will happen is
$p (0.9)^2 (0.04) =  .0324 p$.  The probability that the event won't happen but all three say it will happen is $(1-p)(0.1)^2 (0.96) = .0096 (1-p)$.  So the conditional probability that the event will happen, given that all three say it will happen, is
$$ \dfrac{.0324 p}{.0324 p + .0096 (1-p)}$$
In this case, I happen to think that $p$ is very small, so this is still very 
small: it is approximately $3.375 p$.  
