Question

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)

Answer 1

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