random variables

Question

Consider two independent random variables $X$ and $Y$ with identical distributions.The variables $X$ and $Y$ take value $0, 1$ and $2$ with probabilities $\frac{1}{2},\frac{1}{4}$ $and$  $\frac{1}{4}$ respectively. What is conditional probability $P\left [ \frac{X+Y=2}{X-Y=0} \right ]?$

Comments

Is below approach correct:

There will be 6 (x,y) pair  possible , (0,0),(0,1),(0,2),(1,0),(1,1),(1,2),(2,1),(2,2),(2,3).

So given that x-y =0 p(x+y=2) will be 1/6 as only (1,1) is the favourable case out of total 6

Answer 1

Given X and Y random variables with identical distribution.

$P(X=0)=P(Y=0)=\frac{1}{2}$

$P(X=1)=P(Y=1)=\frac{1}{4}$

$P(X=2)=P(Y=2)=\frac{1}{4}$

Let $A= X+Y=2$

and $B=X-Y=0$

$P\left(\frac{X+Y=2}{X-Y=0}\right)=\frac{P(A∩B)}{P(B)}$

Event $P(A∩B)$ happen when $X+Y=2$ and $X-Y=0$.

this is satisfied only when $X=1$ and $Y=1$

$P(A∩B)=\frac{1}{4} . \frac{1}{4}  = \frac{1}{16}$

and $P(B)$ happen when $X-Y=0$. it occurs when $X=Y$

            i.e. $X=0$ or   $Y=0$

                 $X=1$  or   $Y=1$

                 $X=2$  or   $Y=2$                                 

$P(B)=\frac{1}{2}. \frac{1}{2} + \frac{1}{4}. \frac{1}{4} + \frac{1}{4}. \frac{1}{4}$

$=\frac{6}{16}$

Now 

$P\left(\frac{X+Y=2}{X-Y=0}\right)=\frac{P(A∩B)}{P(B)}=\frac{\frac{\frac{1}{16}}{6}}{16}$

=$\frac{1}{6}$

Comments

Tq 4 ur detailed explanation

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If $A$ and $B$ are independent events then $P(A\cap B)=P(A).P(B)$

                                  $P\left (\frac{A}{B} \right)=\frac{P(A\cap B)}{P(B)}$

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on the above solution  7th line from above is

P((X+Y-2)/P(X-Y-0)) =   P (A  intersection  B)/ P(B)    plz explain ??