Solution to Exercise 5.02.

  (a) The assumed independence gives

$$p_{XY} (x,y) \; = \; p_X (x) p_Y (y) .$$

The results are shown in Table 2.

 

 

1 2 3 4

.300 .050 .025 .025 .100

1 .180 .030 .015 .015 .060

2 .120 .020 .010 .010 .040

(b) Adding the values from the table:

$$P[X \le 1 \, \& \, Y \le 1 ] \; = \; p_{XY} (0,0) + p_{XY} (0,1) + p_{XY} (1,0) + p_{XY} (1,1)$$

$$\; = \; .30 + .05 + .18 + .03 \; = \; .56$$

Computing the individual probabilities:

$$P[X \le 1 ] \; = \; .5 + .3 \; = \; .8 ,$$

$$P[Y \le 1 ] \; = \; .6 + .1 \; = \; .7 ,$$

and of course

$$P[X \le 1 \, \& \, Y \le 1 ] \; = \; P[X \le 1 ] * P[ Y \le 1 ] \; = \; .8 * .7 = .56 .$$

(c)

$$P[X + Y = 0 ] \; = \; P[X = 0 \, \& \, Y = 0 ] \; = \; p_{XY} (0,0) \; = \; .3 .$$

(d)

$$P[ X + Y \le 1 ] \; = \; p_{XY} (0,0) + p_{XY} (0,1) + p_{XY} (1,0)$$

$$\; = \; .30 + .05 + .18 \; = \; .53$$