what is the conditional probability $P(X+Y=2|X-Y=0)$? Consider two independent random variables $X$ and $Y$ with identical distributions The variables takes values $0,1, 2$ with probabilities $\frac12,\frac14,\frac14$. what is the conditional  probability 
$P(X+Y=2|X-Y=0)$?
 A: Answer is 1/6


*

*${(X+Y=2)}\cap{(X-Y=0)} = (X=1) \cap (Y=1)$
reason is : $(X-Y=0)\Longleftrightarrow (X=Y)$...


so : $P((X+Y=2) \cap (X-Y=0))=P(X=1) \times P(Y=1)=(1/4)\times(1/4)=1/16$ (X and Y  being independant)


*

*$P(X-Y=0)=P(X=Y)=P(((X=0) \cap (Y=0)) \cup ((X=1) \cap(Y=1)) \cup((X=2) \cap(Y=2)) = P((X=0) \cap (Y=0))+P((X=1) \cap(Y=1))+P((X=2) \cap(Y=2))=(1/2)\times (1/2)+(1/4)\times (1/4)+(1/4)\times (1/4) = 3/8 $


Finally : $P(X+Y=2/X-Y=0)=\frac{P(X=1)\times P(Y=1)}{P(X=Y)}=\frac{(1/16)}{(3/8)}=1/6$
A: By definition $$P(X+Y=2 | X-Y = 0)= \frac{P(X+Y=2,X=Y)}{P(X=Y)} = \frac{P(X=Y=1)}{P(X=Y)} = \frac{1}{16 \cdot P(X=Y)}.$$ Now try to find $P(X=Y)$.
A: The table of probabilities is (using a common denominator)
X   Y: 0     1     2
-     ----  ----  ---- 
0     4/16  2/16  2/16
1     2/16  1/16  1/16 
2     2/16  1/16  1/16

$X-Y=0$ is the top left to bottom right diagonal.  $X+Y=2$ is the top right to bottom left  diagonal. 
So $P(X+Y=2|X−Y=0) = \dfrac{1/16}{4/16+1/16+1/16} = \dfrac16$.
