Suppose the random variables X, Y, and Z have the following joint probability distribution.

x y z f (x, y, z)

1 1 1 0.05

1 1 2 0.10

1 2 1 0.15

1 2 2 0.20

2 1 1 0.20

2 1 2 0.15

2 2 1 0.10

2 2 2 0.05

1. Determine the following:

a. P (X=2).

b. P (X=1, Y=2).

c. P (Z<1.5).

d. P (X=1 or Z=2)

Step-by-step explanation:

We are given the joint probability distribution of the random variables. Then we should just add the values given, accroding to the case we are asked about.

a. P (X=2) = 0.2 (2 1 1) + 0.15 (2 1 2) + 0.1 (2 2 1) + 0.05 (2 2 2) = 0.5 (We need to add all the values for which X = 2, the values of Y, Z are irrelevant)

b. P (X=1, Y=2) = 0.15+0.2 = 0.35

c. P (Z<1.5) = P (Z=1) = 0.05+0.15+0.2+0.1=0.5 (This means, having Z=1)

d. P (X=1 or Z=2) = P (X=1) + P (Z=2) - P (X=1 and Z=2) (This is using the properties of probability of the union of two events).

P (X=1) = 0.05+0.1+0.15+0.2 = 0.5

P (Z=2) = 0.1+0.2+0.15+0.5 = 0.5

P (X=1 and Z=2) = 0.1+0.2 = 0.3

P (X=1 or Z=2) = 0.5+0.5-0.3 = 0.7