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Given the probability distributions as follows

$P(b_1|a_0,c_0)=p$

$P(b_1|a_1,c_0)=o$

$P(b_1|a_0,c_1)=n$

$P(b_1|a_1,c_1)=m$

$P(a_1|c_1)=x$

$P(a_1|c_0)=y$

$P(c_1)=r$

I need to find $\dfrac{P(b_1|a_0)}{P(a_0)}$

My attempt:

$\dfrac{P(b_1|a_0)}{P(a_0)}=\dfrac{P(b_1|a_0)}{P(a_0,b_0)+P(a_0,b_1)}=\dfrac{P(a_0,b_1,c_1)+P(a_0,b_1,c_0)}{P(a_0,b_0,c_0)+P(a_0,b_0,c_1)+P(a_0,b_1,c_0)+P(a_0,b_1,c_1)}$

I was able to solve the numerator using the given probabilities and the formula $P(a,b,c)=P(b|a,c)\cdot P(a|c)\cdot P(c)$ but I cannot figure out how to solve the denominator part

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  • $\begingroup$ How are $x_1$ and $z_1$ related to $a,b,c$? $\endgroup$
    – gunes
    Commented Sep 19, 2020 at 23:02

1 Answer 1

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Although you haven't explicitly stated, it seems that there are three RVs $a,b,c$ with binary values, e.g. $a_1$ meaning $a=1$.

Given the last three probabilities, i.e. $P(a_1|c_1), P(a_1|c_0), P(c_1)$ you can find the complete joint distribution of $P(a,c)$. And, when you multiply the first four probabilities with corresponding joint distributions of $a$ and $c$, you can find the complete joint distribution of $P(a,b,c)$, from which you can find any probability you like.

For example: $$P(a_0,b_0,c_0)=P(b_0|a_0,c_0)P(a_0,c_0)=(1-p)P(a_0|c_0)P(c_0)=(1-p)(1-y)(1-r)$$

Besides, your numerator is incorrect, it should have been the following: $$P(b_1|a_0)=P(c_1,b_1|a_0)+P(c_0,b_1|a_0)$$

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