Question
question 01 When an automobile is stopped by a roving safety patrol, each tire is checked for tire wear, and each headlight is checked to
question 01
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. LetXdenote the number of headlights that need adjustment, and letYdenote 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) = 0.2, pY(1) = 0.1, pY(2) = pY(3) = 0.05, pY(4) = 0.6, 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) = Does P(X 1 and Y 1) equal the product P(X 1) P(Y 1)?YesNo (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) =
question 02
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.
y | ||||
p(x, y) | 0 | 1 | 2 | |
x | 0 | 0.10 | 0.05 | 0.02 |
1 | 0.06 | 0.20 | 0.08 | |
2 | 0.06 | 0.14 | 0.29 |
(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 {X 0 and Y 0}.One hose is in use on both islands.One hose is in use on one island. At most one hose is in use at both islands.At least 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 | 1 | 2 |
---|---|---|---|
pX(x) |
Compute the marginal pmf of Y.
y | 0 | 1 | 2 |
---|---|---|---|
pY(y) |
Using pX(x), what is P(X 1)?
P(X 1) =
(e)Are X and Y independent rv's? Explain.X and Y are independent because P(x,y) pX(x) pY(y).X and Y are not independent because P(x,y) pX(x) pY(y). X and Y are not independent because P(x,y) = pX(x) pY(y).X and Y are independent because P(x,y) = pX(x) pY(y).
question 03 The joint probability distribution of the number X of cars and the number Y of buses per signal cycle at a proposed left-turn lane is displayed in the accompanying joint probability table.
y | ||||
p(x, y) | 0 | 1 | 2 | |
x | 0 | 0.010 | 0.025 | 0.015 |
1 | 0.020 | 0.050 | 0.030 | |
2 | 0.050 | 0.125 | 0.075 | |
3 | 0.060 | 0.150 | 0.090 | |
4 | 0.040 | 0.100 | 0.060 | |
5 | 0.020 | 0.050 | 0.030 |
(a) What is the probability that there is exactly one car and exactly one bus during a cycle? (b) What is the probability that there is at most one car and at most one bus during a cycle? (c) What is the probability that there is exactly one car during a cycle? Exactly one bus?
P(exactly one car) | = |
P(exactly one bus) | = |
(d) Suppose the left-turn lane is to have a capacity of five cars and one bus is equivalent to three cars. What is the probability of an overflow during a cycle? (e) Are X and Y independent rv's? Explain. Yes, because p(x, y) = pX(x) pY(y).Yes, because p(x, y) pX(x) pY(y). No, because p(x, y) = pX(x) pY(y).No, because p(x, y) pX(x) pY(y).
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