Need help with the last section please 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 ofdefective tires.(a) If X and Y are independent with p,Px(0)0.5, Px(1)= 0.3, Px(2)= 0.2, and p(0) = 0.1, p(1) = 0.6, Py(2) = Py(3) = 0.05, Py(4)= 0.2, display the joint pmf of (X, Y) in a joint probability table.yp(x, y)12340.050.30.0250.0250.110.030.180.0150.0150.0620.020.120.010.010.04(b) Compute P(X < 1 and Y < 1) from the joint probability table.P(X < 1 and Y < 1) = 0.56Does P(X < 1 and Y < 1) equal the product P(X < 1) · P(Y < 1)?O YesO No(c) What is P(X + Y = 0) (the probability of no violations)?P(X + Y = 0) = 0.05(d) Compute P(X + Y < 1).P(X + Y < 1) =

an automobile is stopped by a roving safety patrol, each tire is checked for tire wear, and each headlight tive tires. 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 (a) If X and Y are independent with p(0) = 0.5, Py(1) = 0.3, P(2) = 0.2, and py(0) = 0.1, py(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) 1 4 0 0.05 0.03 0.3 0.025 V 0.025 0.1 1 0.18 0.015 V 0.015 0.06 0.02 0.12 0.01 0.01 0.04 (b) Compute P(X s1 and Ys 1) from the joint probability table. P(X S1 and Y s 1) = 0.56 Does P(X S1 and Ys 1) equal the product P(X s 1) · P(Y S 1)? O Yes O No (c) What is P(X + Y = 0) (the probability of no violations)? (X+x) P(X + Y = 0) = 0.05 (d) Compute P(X + YS 1). P(X + Ys 1) -

an automobile is stopped by a roving safety patrol, each tire is checked for tire wear, and each headlight tive tires. 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 (a) If X and Y are independent with p(0) = 0.5, Py(1) = 0.3, P(2) = 0.2, and py(0) = 0.1, py(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) 1 4 0 0.05 0.03 0.3 0.025 V 0.025 0.1 1 0.18 0.015 V 0.015 0.06 0.02 0.12 0.01 0.01 0.04 (b) Compute P(X s1 and Ys 1) from the joint probability table. P(X S1 and Y s 1) = 0.56 Does P(X S1 and Ys 1) equal the product P(X s 1) · P(Y S 1)? O Yes O No (c) What is P(X + Y = 0) (the probability of no violations)? (X+x) P(X + Y = 0) = 0.05 (d) Compute P(X + YS 1). P(X + Ys 1) -

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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