Answer to Question #128813 in Statistics and Probability for Rachel

Question

Need the answers fast.

The joint probability mass function of (X,Y) is p(x, y) = k(2x + 3y) ; x = 0,1,2; y = 1,2,3. (i)
Find the marginal distributions. (ii) Find P(X = xi / Y = 2) (iii) Find P[X +Y > 3] .

Answer 1

$$p(x, y) = k(2x + 3y)$$

for $X=0,1,2$ and $Y = 1,2,3$

Solutions i)

Marginal distribution of X

$$p(x) =\sum_{\ {\forall Y\\}} p(x,y)$$ $$= \sum_{\ {1<y<3\\}} k(2x + 3y)$$

$$= k(2x + 3) + k(2x + 6) + k(2x + 9)$$

$$=6kx + 18k$$

Answer: p(x) = 6kx + 18k

Marginal distribution of Y

$$p(y) =\sum_{\ {\forall X \\}} p(x,y)$$

$$= \sum_{\ {0<x<2\\}} k(2x + 3y)$$

$$= k(3y) + k(2 + 3y) + k(4 + 3y)$$

$$=9ky + 6k$$

Answer: p(y) = 9ky + 6k

$$\sum_{\ {\forall X \\}} p(x) = \sum_{\ {\forall Y \\}} p(Y) = 1$$

$$\sum_{\ {\forall X \\}} p(x) = \sum_{\ {0<x<2 \\}} 6kx + 18k$$

$$1 = (18k) + (6k + 18k) + (12k + 18k)$$

$$1 = 72k$$

$$=\frac{1}{72}$$

$$p(x,y) = \frac{2x+3y}{72}$$

$$p(x) = \frac{x}{12} + \frac{1}{4}$$

$$p(y) = \frac{y}{8} + \frac{1}{12}$$

Solution ii) P(X = x / Y = 2)

$$p(x,2) = \frac{2x+6}{72}$$

Answer: ( 2 x + 6 ) / 72

Soultion iii) P[X +Y > 3]

$$P[X +Y > 3] = p(X=2 , Y=2) or P(X=2 , Y=3) or P(X=1 , Y=3)$$

$$=\frac{4+6}{72} + \frac{4+9}{72} + \frac{2+9}{72} = \frac{17}{36}$$

Answer: 17 / 36