The joint probability mass function of x and y is given by p(1,1)=0.3p(2,1)=0.1p(3,1)=0.1p(1,2 )=0.05p(2,2)=0p(3,2)=0.05p(1,3)=0.0 5p(2,3)=0.1p(3,3)=0.25 (a) compute the conditional mass function of y given x=2: p(y=1|x=2)= 0.5 p(y=2|x=2)= 0 p(y=3|x=2)= 0.5 (b) are x and y independent? (enter yes or no) no (c) compute the following probabilities: p(x+y> 3)= p(xy=4)= p(xy> 2)= note: you can earn partial credit on this problem. you have attempte

a)

P(Y=1|X=2) = p(2,1) = 0.1

P(Y=2|X=2) = p(2,2) = 0

P(Y=3|X=2) = p(3,2) = 0.1

b)

NO

c)

P(X+Y>3) = 0.55

P(XY=4) = 0

P(XY>2) = 0.55

Step-by-step explanation:

The joint probability table is given in the picture attached

a)

P(Y=1|X=2) = p(2,1) = 0.1

P(Y=2|X=2) = p(2,2) = 0

P(Y=3|X=2) = p(3,2) = 0.1

b)

Two events A and B are independent if, and only if,

P(A|B) = P(A) and P(B|A) = P(B)

X and Y are not independent since

P(Y=1|X=2) = p(2,1) = 0.1

whereas

P(Y=1) = 0.5

so we can see

P(Y=1|X=2)≠ P(Y=1)

c)

P(X+Y>3) = p(1,3)+p(2,2)+p(2,3)+p(3,1)+p(3,2)+p(3,3)=

0.05+0+0.1+0.1+0.05+0.25 = 0.55

P(XY=4) = p(2,2) = 0

P(XY>2) = p(1,3)+p(2,2)+p(2,3)+p(3,1)+p(3,2)+p(3,3) = 0.55