Given the joint distribution P(1,1) and P(0,0) of P(X,Y), how do i find P(0,1) and P(1,0)?

Question

Given the joint distribution $P(X,Y)$ where $P(0,0) = 0.25$ and $P(1,1)= 0.6$.

$X$ and $Y$ can only take the values of $\{0,1\}$. Also it is given that $X \sim \textrm{Bern(p)}$. I've been searching for countless hours trying to find formulas to on how will $X \sim \textrm{Bern(p)}$ or the other given joint distribution help me find the others.

Edit: I just realized that I made a mistake. Instead of typing them as conditional probabilities, i typed them as joint probabilities.

Answer 1

You can make a table

$$\begin{array}{c|c|c|c} X \backslash Y&0&1& \sum\\ \hline 0 &0.25 & & 1-p \\ \hline 1 & & 0.6 & p\\ \hline \sum & & & 1 \end{array}$$

Since $X$ is bernoulli distributed we have two states, $0$ and $1$. The probability for state $1$ then is $p$. Thus $P(X=1)=p$ and $P(X=0)=1-p$

For a specific value of p the remaining values can be calculated with simples algebra.

Answer 2

Hint: the law of total probability tells us

$$P(X=1)=P(X=1,Y=1)+P(X=1,Y=0).$$

Use this to solve $P(X=1,Y=0)$. Finally, use the fact

$$1=P(X=1,Y=1)+P(X=1,Y=0)+P(X=0,Y=1)+P(X=0,Y=0)$$

to solve $P(X=0,Y=1)$.