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Bayes' law is not needed. Let $M_i = \text{" the data is in memory } m_i"$. We're given that $$\Bbb{P}(M_2) = 0.6$$ and, since the data is not in $m_3$ if and only if it is in $m1$ or $m2$, we also have the given value $$\Bbb{P}(\overline{M_3}) = \Bbb{P}(M_1 \cup M_2) = 0.76$$ Note also that since the data is in only one of the three memories, ...

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You have a prior distribution $\pi(q)$, a likelihood function $p(x|q)$ and two observations $x_1=1$ and $x_2=1$. So your posterior distribution for $q$ is $$\pi(q|x_1,x_2)=\frac{\pi(q)p(x_1|q)p(x_2|q)}{\displaystyle \int_0^1 \pi(r)p(x_1|r)p(x_2|r)\; dr}$$ in effect taking the product of the prior and the likelihoods and then turning this into a probability ...

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Consider the following analogous problem. Suppose that a certain country issues ID cards to all nationals aged 20, and that the birth rate is the same for men and women. You work for the department that handles applications for ID cards, and have in your hands an envelope containing a randomly selected application form for a new card. John offers you a ...

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OP here, in case anyone has a similar question, I believe this can be done with parameter estimation through the use of "expectation maximisation algorithm", see here

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Your first calculation is correct. $P(A)=.0545$. For your second calculation, in fact, $\Pr(A\cap E)\ne 0$ (it equals $(.005)(.95)=.00475$.

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The usual method to solve such questions for Bayesian networks is to use the Chain Rule for probability and use the rules regarding independence and conditional independence implied by the lines joining the nodes. Firstly, $P(\lnot A, C, D) = P(\lnot A, C, D, B) + P(\lnot A, C, D, \lnot B)$, since either $B$ must occur or not occur. We can calculate each ...

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Other than the normal rules on conditional probability, the key thing specific to working with Bayesian networks (BN) is the rules regarding independence and conditional independence of the events represented by the nodes. Without using the independence inferred by the particular network you can only get so far. There are lots of notes around if you google ...

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