Homework SourseThe joint probability distribution of the number X of cars a
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.
p(x, y)
(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?
(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).
| y | ||||
| p(x, y) | 0 | 1 | 2 | |
| x | 0 | 0.015 | 0.025 | 0.010 |
| 1 | 0.030 | 0.050 | 0.020 | |
| 2 | 0.075 | 0.125 | 0.050 | |
| 3 | 0.090 | 0.150 | 0.060 | |
| 4 | 0.060 | 0.100 | 0.040 | |
| 5 | 0.030 | 0.050 | 0.020 |
Solution
a) P(X=1, y =1) = 0.050
b) P(X<=1, y<=1) = P(0,0)+P(0,1)+P(1,0)+P(1,1)
= 0.015+0.025+0.030+0.050
= 0.120
c) P(X=1) = 0.030+0.050+0.020 = 0.100
P(y=1) = 0.50
d) P(x>5) =P(y = 2)+P(y=3)+P(y=4)+P(y=5)
= 0.85
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Let us consider P(X=0, y=0) = 0.015
P(X=0) = 0.05 and P(Y=0) = 0.3
Seeing the table we find this holds good for all x and y
Hence x and y are independent.
Yes, because p(x, y) = pX(x) · pY(y)
P(X)P(Y) = 0.015
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