(a) If X and Y are independent with p(0) = 0.5, py(1) = 0.3, p(2) = 0.2, and p,(0) = 0.1, p,(1) = 0.6, p(2) = Py(3) = 0.05, Py(4) = 0.2, display the joint pmf of (X, Y) in a joint probability table. p(x, y) 3 2 4 2 (b) Compute P(X < 1 and Y s 1) from the joint probability table. P(X s1 and Y S 1) =| Does P(X s 1 and Ys 1) equal the product P(X < 1) · P(Y < 1)? O Yes O No (c) What is P(X + Y = 0) (the probability of no violations)? P(X + Y = 0) =
(a) If X and Y are independent with p(0) = 0.5, py(1) = 0.3, p(2) = 0.2, and p,(0) = 0.1, p,(1) = 0.6, p(2) = Py(3) = 0.05, Py(4) = 0.2, display the joint pmf of (X, Y) in a joint probability table. p(x, y) 3 2 4 2 (b) Compute P(X < 1 and Y s 1) from the joint probability table. P(X s1 and Y S 1) =| Does P(X s 1 and Ys 1) equal the product P(X < 1) · P(Y < 1)? O Yes O No (c) What is P(X + Y = 0) (the probability of no violations)? P(X + Y = 0) =
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
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Transcribed Image Text:When an automobile is stopped by a roving safety patrol, each tire is checked for tire wear, and each headlight is checked to see whether it is properly aimed. Let X denote the number of headlights
that need adjustment, and let Y denote the number of defective tires.
(a) If X and Y are independent with py(0) = 0.5, Px(1) = 0.3, px(2) = 0.2, and py(0) = 0.1, p(1) = 0.6, p(2) = Py(3) = 0.05, p(4) = 0.2, display the joint pmf of (X, Y) in a joint
probability table.
y
p(x, y)
1
4
1
2
(b) Compute P(X < 1 and Y < 1) from the joint probability table.
P(X < 1 and Y< 1) =
Does P(X < 1 and Y < 1) equal the product P(X < 1) · P(Y < 1)?
O Yes
O No
(c) What is P(X + Y = 0) (the probability of no violations)?
P(X + Y = 0) =
(d) Compute P(X + Y < 1).
P(X + Y < 1) =
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