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 pX(0) = 0.5, pX(1) = 0.3, pX(2) = 0.2, and pY(0) = 0.6, pY(1) = 0.1, pY(2) = pY(3) = 0.05, pY(4) = 0.2, display the joint pmf of (X, Y) in a joint probability table. y p(x, y) 0 1 2 3 4 x 0 1 2 (b) Compute P(X ≤ 1 and Y ≤ 1) from the joint probability table. P(X ≤ 1 and Y ≤ 1) = (c) What is P(X + Y = 0) (the probability of no violations)? P(X + Y = 0) =
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 pX(0) = 0.5, pX(1) = 0.3, pX(2) = 0.2, and pY(0) = 0.6, pY(1) = 0.1, pY(2) = pY(3) = 0.05, pY(4) = 0.2, display the joint pmf of (X, Y) in a joint probability table. y p(x, y) 0 1 2 3 4 x 0 1 2 (b) Compute P(X ≤ 1 and Y ≤ 1) from the joint probability table. P(X ≤ 1 and Y ≤ 1) = (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
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Problem 1P
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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 pX(0) = 0.5, pX(1) = 0.3, pX(2) = 0.2, and pY(0) = 0.6, pY(1) = 0.1, pY(2) = pY(3) = 0.05, pY(4) = 0.2, display the joint pmf of (X, Y) in a joint probability table.
(b) Compute P(X ≤ 1 and Y ≤ 1) from the joint probability table.
P(X ≤ 1 and Y ≤ 1) =
y | ||||||
p(x, y)
|
0 | 1 | 2 | 3 | 4 | |
x | 0 | |||||
1 | ||||||
2 |
(b) Compute P(X ≤ 1 and Y ≤ 1) from the joint probability table.
P(X ≤ 1 and Y ≤ 1) =
(c) What is P(X + Y = 0) (the probability of no violations)?
P(X + Y = 0) =
P(X + Y = 0) =
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