00:01
Here the two variables x and y are two discrete random variables.
00:07
Now we have to find the probability mass function with respect to the conditional property.
00:12
X and y values.
00:14
So x takes the values 1, 2, 3 and p of y equal to y we have to find and y takes the values 1 2 and p of x equal to x.
00:24
So here the corresponding probabilities are 0 .2 multiplied by 1 which is 0 .0 .0 .0.
00:31
Then here 0 .04, 0 .56, 0 .08, 0 .06 and 0 .24.
00:41
And the total is 0 .62, 0 .38 .8.
00:48
And here, 0 .10.
00:53
So, here the total is 1 .62, 0 .38.
01:01
And here, the total is 1.
01:02
P of x equal to x comma y equal to y now e of x is equal to summation x times p of x which is equal to 1 multiplied by 0 .1 plus 2 multiplied by 0 .1 plus 3 multiplied by 0 .8 so which is equal to 2 .7 that is the expected value next we have to find e of 1 by y which is equal to 1 by y p of y equal to y summation of this one.
01:34
So which is equal to 1 multiplied by 0 .62 plus 1 by 2 multiplied by 0 .38, which is equal to 0 .62 plus 0 .19, which is equal to 0 .81...