The joint probability distribution of the number X of cars...

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. (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 cvcle? 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 cucle? (e) Are $X$ and $Y$ independent rvs? Explain.

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.
(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 cvcle? 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 cucle?
(e) Are $X$ and $Y$ independent rvs? Explain.

Key Concepts

Transformation of Random Variables

Transformation of random variables involves modifying the original variables by applying a function, such as a conversion or scaling factor, to obtain a new random variable. This is particularly useful in capacity or equivalence problems where different units (e.g., converting buses to an equivalent number of cars) must be compared or aggregated to assess thresholds or limits.

Independence of Random Variables

Random variables are considered independent if the joint probability of any pair of outcomes is equal to the product of their individual marginal probabilities. Examining the independence involves checking whether the structure of the joint probability distribution factorizes into the marginals, which simplifies analysis and calculations in probability theory.

Marginal Probability Distribution

Marginal probability distributions are obtained from the joint distribution by summing or integrating over one or more of the variables. This gives the probability distribution for a single variable irrespective of the values of the other variables, which is crucial for analyzing individual behaviors within a joint framework.

Joint Probability Distribution

A joint probability distribution specifies the probability that each pair of outcomes (or combination of outcomes) for two or more random variables occurs. It is typically represented in a table or mathematical function and provides a complete description of the relationship between the variables, including any dependencies or correlations that may exist.

Compound Event Probability

Compound event probability involves determining the likelihood of events that are formulated from multiple conditions, such as combining outcomes from two different random variables. This may include events defined by intersections (e.g., both conditions occurring simultaneously) or unions (e.g., at least one of the conditions occurring), and is computed directly from the joint or marginal probabilities.

Transcript

00:01 We're told that 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 given in the accompanying probability table.

00:15 This is an exercise 9 of section 4 .1.

00:23 In part a, we're asked to find the probability that there is exactly one car and exactly one bus during a cycle.

00:31 So the probability of exactly one car and exactly one bus.

00:40 This is the same as p of 1 -1, which looking at our table, it's easy to find this is .03.

00:55 In part b, we're asked to find the probability that there is at most one car and at most one bus during a cycle.

01:04 So this is the same as the probability that x is less than or equal to 1 and that y is less than or equal to 1.

01:13 So this is, well, p of 0 is an option, plus p of 01, plus p of 1 0, plus p of 1, 1.

01:26 And these are all the options.

01:29 And adding these together, after referring to the table to find these values, we get 0 .120.

01:44 In part c, we're asked to find that probability that there is exactly one card during a cycle, and then we're asked to find the probability that there is exactly one bus.

01:56 So if there's exactly one car during a cycle, this is the same as the probability that x is equal to 1, and this is p of 1 -0 plus p of 1 -2, we're adding overall possible values of y, except i should have in p of 1 -1 here, plus p of 1 -2, and adding these together, this is equal to 0 -10, and to find probability that there is exactly one bus during a cycle.

02:49 This is the probability that y equals 1, which is the same as summing across or down the column for y equals 1, p of 0 1 plus p of 1 plus p of 1 1 plus p of 3 1 plus p of 4 1 plus p of 5 1, and adding all these together, we get .300.

03:41 Finally, in part d, unless isn't finally, we're told that the left turn lane is going to have the capacity of five cars, and that one bus is equivalent to three cars.

03:54 We're asked to find the probability of an overflow during a cycle.

04:03 So the probability of an overflow, well, when would an overflow occur? this is when we have that the number of cars plus, we'll recall that cars, three cars are equivalent to one bus.

04:35 So we have the total number of cars, which is x plus three times the number of buses y.

04:57 And since it only has a capacity of five cars, this is the probability that this x plus 3y is greater than five.

05:04 This is of course the same as 1 minus the probability that x plus 3y is less than or equal to 5.

05:17 This is easier to find, which is equal to 1 minus the probability that, well, we could have, thinking of examples, say x and y are both 0, this will work.

05:44 You could also have that x is 0 and y is 1.

05:48 You could also have that x is 1 and y is 0.

05:54 We could also have that x is 1 and y is 1.

06:02 This would give us 4.

06:08 We could also have that x is 2, y is 0.

06:20 We could also have that x is 2 and y is 1.

06:25 This will give us total of 5.

06:30 We could also have that x is 3 and y is 0.

06:45 And we could have that x is 4 and y is 0...