-/6 points DevoreStat95E.007. My Notes Ask Your Teacher The joint probability di

Question

-/6 points DevoreStat95E.007. My Notes Ask Your Teacher 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) 0 0 0.015 0.025 0.010o 1 0.030 0.050 0.020 2 0.075 0.125 0.050 0.090 0.150 0.060 4 0.060 0.100 0.040 5 0.030 0.050 0.020 (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 O Yes, because (x, y) = pdx) , py(y) O Yes, because p(x, ) Px) py) No, because p(x, )x)P) O No, because (x, y) * px(x) , py(y).

Explanation / Answer

ANSWERS:

a.) The probability that there is exactly one car and exactly one bus during the cycle:

P(1,1) = 0.050

b.) Probability that there is at most one car and at most one bus during a cycle?

=P(0,0) + P(0,1) +P(1,0) +P(1,1)

=0.015 + 0.025 + 0.030 + 0.050

= 0.12

c.) P(exactly one car) =0.030+0.050+0.020 = 0.1

P(exactly one bus) =0.025 +0.050+0.125+0.150+0.100+0.050 = 0.5

e.) Are X and Y indpedent rv`s?

P(0,0)= 0.015

P(X=0)= 0.015+0.025+0.010= 0.05

P(Y=0) = 0.015+0.030+0.075+0.090+0.060+0.030=0.3

P(X=0)*P(Y=0) =0.3*.0.05 =0.015

P(0,0) = P(X=0)* P(Y=0)

Hence X and Y are independent.

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