Compute $P(A=0, B=0)$ given random variables $X$ and $Y$

Question

Given random variables $X$ and $Y$ with a $Ber(\frac{1}{2})$ Distribution, define random variable $A = X + Y$ and $B = |X - Y|$

Compute

1. $P(A = 0, B = 0)$
2. $P(A = 1)$

I'm not quite sure how to approach this, I've tried to individually add up the probabilities as in $P(X = 0) = 0.5$ and so forth but that doesn't seem to work out.

Would appreciate any help, thanks in advance!

Answer 1

I'll be treating the case where $X$ and $Y$ are independent.

Since $X\in \{0,1\}$ and $Y\in \{0,1\}$, then $$X+Y=0 \text{ and }|X-Y|=0 \implies X=Y=0. $$ Then $$\mathbb P(A=0,B=0)=\mathbb P(X=0,Y=0),$$ and it stops here.

If however $X$ and $Y$ are independent, then $$\mathbb P(X=0,Y=0)=\mathbb P(X=0)\times\mathbb P(Y=0)=\frac{1}{2}\times\frac{1}{2}=\frac{1}{4}.$$ Same applies on the second one (if we assume the independence) \begin{align*} \mathbb P(A=1)&=\mathbb P(X+Y=1)\\ &=\mathbb P\left((X=0,Y=1)\vee(X=1,Y=0)\right)\\ &=\mathbb P(X=1,Y=0)+\mathbb P(X=0,Y=1)-\underbrace{\mathbb P((X=1,Y=0)\cap(X=0,Y=1))}_{\emptyset}\\ &=\mathbb P(X=1,Y=0)+\mathbb P(X=0,Y=1) \leftarrow\text{It stops here }\\ &=\left(\mathbb P(X=1)\times\mathbb P(Y=0)\right)+\left(\mathbb P(X=0)\times\mathbb P(Y=1)\right)\\ &=\left(\frac{1}{2}\times\frac{1}{2}\right)+\left(\frac{1}{2}\times\frac{1}{2}\right)\\ &=\frac{1}{2} \end{align*}'