Question

3.6. Problem*. (Section 9.1) Currently, the House of Commons in Britain includes 288! Conservative party members, 245 Labour party members, and 117 members from other political parties. Suppose two members of the House of Commons are randomly selected. Let, X Y Z = number of Conservative party members selected, = number of Labour party members selected = number of members selected that are not Conservative or Labour party members (a) Construct a sequence of three tables that describe the joint probability mass function of X, Y, and Z. (b) Calculate the probability that X Y > Z. (c) Calculate E[X|Y = 2].

Answer 1

Following is an exhaustive table of all possible options of selections. Individual probabilities are calculated as C [n,r]

X	Y	Z	P
0	0	2	0.032
0	1	1	0.136
0	2	0	0.141
1	0	1	0.160
1	1	0	0.335
2	0	0	0.196

Joint probability tables are as below:

			X
		0	1	2
	0	0.032	0.16	0.196
Y	1	0.136	0.335	0
	2	0.141	0	0
			X
		0	1	2
	0	0.141	0.335	0.196
Z	1	0.136	0.16	0
	2	0.032	0	0
			Y
		0	1	2
	0	0.196	0.335	0.141
Z	1	0.16	0.136	0
	2	0.032	0	0

(b) P (X >= Y >= Z) = P (X=1, Y=1, Z=0) + P(X=2, Y=0, Z=0) = 0.335 + 0.196 = 0.53

(c) E (X / Y=Z) = E[X / Y=Z=0 & Y=Z=1] = 2 x 0.196 + 0 x 0.136 = 0.392