What is the difference between $p(a|b)$ and $p(a)$?

Question

I know that $$p(a|b)=\frac{p(a, b)}{p(b)}$$ And I also know $$p(a, b) = p(a)p(b)$$

So, algebraically, it all seems to me that $$p(a|b)=p(a)$$ I know something is wrong with this situation that I'm thinking about but I don't know where I am wrong.
My problem is that I have Bayesian network like this image. I have the probability distribution for the MotherGene and FatherGene, and I wanna calculate the conditional probability $p(ChildGene=1|MotherGene=0,FatherGene=2)$. So it's gonna be $$ \frac{p(ChildGene=1, MotherGene=0, FatherGene=2)}{p(MotherGene=0, FatherGene=2)}$$ It's exactly like $p(ChildGene=1)$ when I try to calculate and the Bayesian network doesn't affect anything.

Answer 1

For example if eye colour is the gene, let $p(B)$ be the probability that the child's eye colour is blue, and $P(B|MB)$ be the probability that the child's eye colour is blue given that the mother's eye colour is blue.

Suppose the probability that the eye colour is blue is 0.25 but mothers with blue eyes give birth to children with blue eyes 50% of the time then $$ P(B)=0.25\\ P(B|MB)=0.5 $$ Now assuming mothers come from the same population as the children, $P(MB)=0.25$, so $P(B)P(MB)=0.25^2={1\over 16}$ but $P(B|MB)=0.5\neq P(B)P(MB)$.

Answer 2

To give a more detailed explanation -- For two events $a$ and $b$, the quantity $p(a\vert b)$ is called the probability of $a$ given $b$ and should be understood as such. It indeed satisfies the formula $$ p(a\vert b) = \frac{p(a,b)}{p(b)}\quad \hbox{provided }p(b)>0. $$ These events are said to be independent if $p(a,b) = p(a)p(b)$. It sometimes happens that two events are independent while still "interacting with one another in some way" -- so you should not be too alarmed by your observation.

To give a concrete (classical) example, we can toss three fair coins and denote by $a_{ij}$ the event "coin $i$ and $j$ match" ($1\le i<j\le 3$) so that $p(a_{ij}) = 1/2$. Now notice that the events $a_{12},a_{23},a_{13}$ are pairwise independent as $$ p(a_{ij},a_{ik}) = \frac{1}{4} = p(a_{ij})p(a_{ik}), $$ whenever $j\ne k$. However these events are not independent since \begin{align*} p(a_{12},a_{21},a_{13}) = p\left( \hbox{all coins match} \right) = \frac{1}{4}\ne \frac{1}{8} = p(a_{12})p(a_{23})p(a_{13}). \end{align*} Coming back to your point of view of conditional probability, we can note that $p(a_{12}\vert a_{13}) = p(a_{12}) = p(a_{12}\vert a_{23})$. In other words the fact that coin $1$ or coin $2$ matches with coin $3$ does not affect the likelihood of the first two coins to match. However $p(a_{12}\vert a_{13},a_{23}) = 1$. Indeed, if coin $1$ matches with coin $3$ and so do $2$ and $3$ then, we must have coin $1$ and $2$ that match.

I hope this provides an intuition to the fact that events can be independent whilst still "interacting" with one another.