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) -
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 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.
y
p(x, y)
1
2
3
4
0.05
0.3
0.025
0.025
0.1
1
0.03
0.18
0.015
0.015
0.06
2
0.02
0.12
0.01
0.01
0.04
(b) Compute P(X < 1 and Y < 1) from the joint probability table.
P(X < 1 and Y < 1) = 0.56
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) = 0.05
(d) Compute P(X + Y < 1).
P(X + Y < 1) =
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