The joint probability distribution of the number X of cars and the number Y of b

Question

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  

Explanation / Answer

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

p(x, y) 0 1 2      x 0     0.015     0.025     0.010   0.05 1     0.030     0.050     0.020   0.1 2     0.075     0.125     0.050   0.25 3     0.090     0.150     0.060   0.3 4     0.060     0.100     0.040   0.2 5     0.030     0.050     0.020 0.1 0.30 0.50 0.20 1.00

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