Two dependent Bernouilli trials with probability $p$ and $q$

Question

Say you have two Bernouilli trials which have normally probability $p$ and $q$ respectively. If they are independent, the probability of having 1,1 is $p \times q$. Of having 0,0 is $(1-p)\times(1-q)$ , 1,0 = $(p)\times(1-q)$

But now, if both trials are not independent. The probability of seeing 1,1 is probably more than $p \times q$ and of seeing 1,0 is less than $(p)\times(1-q)$.

How have people modeled this previously ?

Answer 1

you need to define the nature of the dependence by introducing conditional probabilities

Answer 2

Let $\Omega:=\{\langle0,0\rangle,\langle0,1\rangle,\langle1,0\rangle,\langle1,1\rangle\}$ and let $\mathcal A:=\wp(\Omega)$.

Then determine the probability $P$ by stating that $P(\{\langle i,j\rangle\})=p_{i,j}$ where $p_{i,j}\geq0$ and $$p_{0,0}+p_{0,1}+p_{1,0}+p_{1,1}=1$$

Note that this equality makes it possible to practicize exactly $3$ parameters as Henry suggests in his comment.

The functions $B_1,B_2:\Omega\rightarrow\{0,1\}$ prescribed by $\langle i,j\rangle\mapsto i$ and $\langle i,j\rangle\mapsto j$ respectively are both Bernouilli-distributed.

This with $P(B_1=1)=p_{1,0}+p_{1,1}$ and $P(B_2=1)=p_{0,1}+p_{1,1}$.