(e) Are X and Y independent rv's? Explain. Ox and Y are independent because P(x, y) = Px(x) · Py(y). OX and Y are not independent because P(x, y) = Px(x) · Py(y). OX and Y are independent because P(x, y) OX and Y are not independent because P(x, y) Px(x) · Py(y). Px(x) ·Py(y). A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let X denote the number of hoses being used on the self-service island at a particular time, and let Y denote the number of hoses on the full-service island in use at that time. The joint pmf of X and Y appears in the accompanying tabulation. p(x, y) 0 1 2 0 0.10 0.05 0.01 x 1 0.06 0.20 0.07 2 0.06 0.14 0.31 (a) What is P(X = 1 and Y = 1)? P(X = 1 and Y = 1) = | (b) Compute P(X ≤ 1 and Y < 1). P(X ≤1 and Y < 1) = (c) Give a word description of the event { X0 and Y + 0 }. One hose is in use on one island. One hose is in use on both islands. At least one hose is in use at both islands. At most one hose is in use at both islands. Compute the probability of this event. P(X +0 and Y 0) = [ (d) Compute the marginal pmf of X. x 0 Px(x) Compute the marginal pmf of Y. y Py(y) 0 Using Px(x), what is P(X ≤ 1)? P(X ≤ 1) = = 1 2 1 2
(e) Are X and Y independent rv's? Explain. Ox and Y are independent because P(x, y) = Px(x) · Py(y). OX and Y are not independent because P(x, y) = Px(x) · Py(y). OX and Y are independent because P(x, y) OX and Y are not independent because P(x, y) Px(x) · Py(y). Px(x) ·Py(y). A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let X denote the number of hoses being used on the self-service island at a particular time, and let Y denote the number of hoses on the full-service island in use at that time. The joint pmf of X and Y appears in the accompanying tabulation. p(x, y) 0 1 2 0 0.10 0.05 0.01 x 1 0.06 0.20 0.07 2 0.06 0.14 0.31 (a) What is P(X = 1 and Y = 1)? P(X = 1 and Y = 1) = | (b) Compute P(X ≤ 1 and Y < 1). P(X ≤1 and Y < 1) = (c) Give a word description of the event { X0 and Y + 0 }. One hose is in use on one island. One hose is in use on both islands. At least one hose is in use at both islands. At most one hose is in use at both islands. Compute the probability of this event. P(X +0 and Y 0) = [ (d) Compute the marginal pmf of X. x 0 Px(x) Compute the marginal pmf of Y. y Py(y) 0 Using Px(x), what is P(X ≤ 1)? P(X ≤ 1) = = 1 2 1 2
Computer Networking: A Top-Down Approach (7th Edition)
7th Edition
ISBN:9780133594140
Author:James Kurose, Keith Ross
Publisher:James Kurose, Keith Ross
Chapter1: Computer Networks And The Internet
Section: Chapter Questions
Problem R1RQ: What is the difference between a host and an end system? List several different types of end...
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Transcribed Image Text:(e) Are X and Y independent rv's? Explain.
Ox and Y are independent because P(x, y) = Px(x) · Py(y).
OX and Y are not independent because P(x, y) = Px(x) · Py(y).
OX and Y are independent because P(x, y)
OX and Y are not independent because P(x, y)
Px(x) · Py(y).
Px(x) ·Py(y).
Transcribed Image Text:A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with
two hoses. Let X denote the number of hoses being used on the self-service island at a particular time, and let Y denote the
number of hoses on the full-service island in use at that time. The joint pmf of X and Y appears in the accompanying
tabulation.
p(x, y)
0 1 2
0 0.10 0.05 0.01
x
1 0.06 0.20 0.07
2 0.06 0.14 0.31
(a) What is P(X = 1 and Y = 1)?
P(X = 1 and Y = 1) = |
(b) Compute P(X ≤ 1 and Y < 1).
P(X ≤1 and Y < 1) =
(c) Give a word description of the event { X0 and Y + 0 }.
One hose is in use on one island.
One hose is in use on both islands.
At least one hose is in use at both islands.
At most one hose is in use at both islands.
Compute the probability of this event.
P(X +0 and Y 0) = [
(d) Compute the marginal pmf of X.
x
0
Px(x)
Compute the marginal pmf of Y.
y
Py(y)
0
Using Px(x), what is P(X ≤ 1)?
P(X ≤ 1) =
=
1
2
1
2
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