A small department has 10 faculty members of whom 3 are assistant professors (aP), 5 are associate professors (AP), and 2 are professors (P). Of these 10 faculty members, 2 are randomly selected to be on a committee. Let X denote the number of aP selected and let Y denote the number of AP selected. X and Y are jointly discrete random variables with mass points (x, y) = {(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (2, 0) Moreover, we can find the probability associated with each of these mass points as follows: ƒ(0,0) = P(X = 0, Y = 0) = P(2 P selected) == 1 (3) f(0, 1) = P(X= 0, Y = 1) = P(1 P, 1 AP selected): = (2) = ==

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A small department has 10 faculty members of whom 3 are assistant professors (aP), 5 are associate
professors (AP), and 2 are professors (P). Of these 10 faculty members, 2 are randomly selected to be
on a committee. Let X denote the number of aP selected and let Y denote the number of AP selected.
X and Y are jointly discrete random variables with mass points (x, y) = {(0,0), (0, 1), (0, 2), (1, 0), (1, 1), (2,0)}.
Moreover, we can find the probability associated with each of these mass points as follows:
f(0,0) = P(X= 0, Y = 0) = P(2 P selected)
f(0, 1) = P(X= 0, Y = 1) = P(1 P, 1 AP selected)
:
= P(X = 2, Y = 0) = P(2 aP selected)
8 = 1/3
Leading to the joint pmf: f(x, y) = P(X = x, Y = y) =
X = Number of aP chosen
0
1
1/45 6/45
10/45 f(1,1)
Y = Number of AP chosen 0
1
2 10/45 0
=
Given this joint pmf,
54. (a) Find f(1, 1) = P(X = 1, Y = 1)
=
2
3/45
0
0
=
(b) Find the marginal distribution of X, fx (x).
(c) Find the marginal distribution of Y, fy (y).
(d) Find the probability that exactly one aP will be chosen.
4/55
-
Transcribed Image Text:A small department has 10 faculty members of whom 3 are assistant professors (aP), 5 are associate professors (AP), and 2 are professors (P). Of these 10 faculty members, 2 are randomly selected to be on a committee. Let X denote the number of aP selected and let Y denote the number of AP selected. X and Y are jointly discrete random variables with mass points (x, y) = {(0,0), (0, 1), (0, 2), (1, 0), (1, 1), (2,0)}. Moreover, we can find the probability associated with each of these mass points as follows: f(0,0) = P(X= 0, Y = 0) = P(2 P selected) f(0, 1) = P(X= 0, Y = 1) = P(1 P, 1 AP selected) : = P(X = 2, Y = 0) = P(2 aP selected) 8 = 1/3 Leading to the joint pmf: f(x, y) = P(X = x, Y = y) = X = Number of aP chosen 0 1 1/45 6/45 10/45 f(1,1) Y = Number of AP chosen 0 1 2 10/45 0 = Given this joint pmf, 54. (a) Find f(1, 1) = P(X = 1, Y = 1) = 2 3/45 0 0 = (b) Find the marginal distribution of X, fx (x). (c) Find the marginal distribution of Y, fy (y). (d) Find the probability that exactly one aP will be chosen. 4/55 -
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