A friend who lives in Los Angeles makes frequent consulting...

A friend who lives in Los Angeles makes frequent consulting trips to Washington, D.C.; 50$\%$ of the time she travels on airline $\# 1,30 \%$ of the time on airline $\# 2,$ and the remaining 20$\%$ of the time on airline $\# 3 .$ For airline $\# 1,$ flights are late into D.C. 30$\%$ of the time and late into L. 10$\%$ of the time. For airline $\# 2,$ these percentages are 25$\%$ and $20 \%,$ whereas for airline $\# 3$ the percentages are 40$\%$ and 25$\%$ . If we learn that on a particular trip she arrived late at exactly one of the two destinations, what are the posterior probabilities of having flown on airlines $\# 1, \# 2,$ and $\# 3 ?$ Assume that the chance of a late arrival in L. A. is unaffected by what happens on the flight to D.C. [Hint: From the tip of each first-generation branch on a tree diagram, draw three second- generation branches labeled, respectively, 0 late, 1 late, and 2 late. $]$

A friend who lives in Los Angeles makes frequent consulting trips to Washington, D.C.; 50$\%$ of the time she travels on airline $\# 1,30 \%$ of the time on airline $\# 2,$ and the remaining 20$\%$ of the
time on airline $\# 3 .$ For airline $\# 1,$ flights are late into D.C. 30$\%$ of the time and late into L. 10$\%$ of the time. For airline $\# 2,$ these percentages are 25$\%$ and $20 \%,$ whereas for airline $\# 3$ the percentages are 40$\%$ and 25$\%$ . If we learn that on a particular trip she arrived late at exactly one of the two destinations, what are the posterior probabilities of having flown on airlines $\# 1, \# 2,$ and $\# 3 ?$ Assume that the chance of a late arrival in L. A. is unaffected by what happens on the flight to D.C. [Hint: From the tip of each first-generation branch on a tree diagram, draw three second-
generation branches labeled, respectively, 0 late, 1 late, and 2 late. $]$

Key Concepts

Probability Tree Diagrams

Probability tree diagrams are visual tools that help in mapping out complex probability problems. They break down processes into branches that represent sequential events and their probabilities, aiding in the organization and computation of joint and conditional probabilities.

Independence of Events

Independence of events means that the occurrence or non-occurrence of one event does not affect the probability of another event. Recognizing independence is key when combining probabilities, as it allows for the direct multiplication of probabilities for independent events.

Bayes Theorem

Bayes Theorem is a fundamental concept in probability that allows us to update prior beliefs based on new evidence. It relates the conditional probability of an event given prior information to the likelihood of observing that evidence under different conditions, making it especially useful in situations where we wish to compute the probability of an underlying cause given an observed outcome.

Conditional Probability

Conditional probability quantifies the likelihood of an event occurring given that another event has already occurred. This concept is core to understanding how probabilities change with additional information and is essential when events are interconnected, whether or not they are independent.

Law of Total Probability

The Law of Total Probability is used to decompose the probability of an event into several disjoint components that cover all possible ways the event can occur. By summing the contributions from all different scenarios, each weighted by its probability, this concept assists in calculating the overall likelihood of an outcome.

Transcript

00:01 All right, so we're given some information about a friend of ours who likes going between los angeles and washington, d .c.

00:07 We're given three airlines, the probability that that friend uses each airline, and the probability that they're late pulling either out of washington c or into los angeles.

00:18 So we have a hint that recommends that we set up a probability tree in a certain way.

00:22 So we're going to set up that probability tree right now.

00:28 All right.

00:29 So let a1 be airline 1, a2, b, airline 2, a2, a2, et cetera.

00:42 So the probability of a1 is 0 .50, probability of a2 is 0 .30, the probability of a3 is 0 .20.

00:50 That's what we're given.

00:52 All right.

00:53 Now i'm going to define a new variable, l0, l1, and l2.

01:00 And what these mean is as follows.

01:02 L0 means that there are no late flights on either end.

01:11 L1 means there's exactly one late flight, whether that be in l .a.

01:15 Or washington, d .c.

01:23 And l2 means both flights are late.

01:31 Now, we're given the statistics that cernan either landing or taking off is late in each city, and we're going to use that to our advantage.

01:41 So we're going to make this like three -pronged thing off the side of each of our branches.

01:55 Right.

01:57 First off, we're going to find l2.

01:59 So we're going to say probability of l2 given a1.

02:04 We're given that they're late on airline 1.

02:07 Flights are late into dc at 30 % of the times and late into la at 10 % of the time.

02:13 So that means the probability of both occurring is 0 .3 times 0 .03 and 0 .03.

02:22 Conversely, the probability of no late flights on a1 is going to be 1 minus the probability that the flights are late in dc, so 0 .7, times 1 minus the probability they're late in los angeles.

02:37 So 0 .9.

02:38 This is 0 .63.

02:41 Now for l1, i'm going to eschew the multiplication for this because these are only three possibilities, which means the last possibility is just going to be one minus all of the probabilities.

02:51 So the other probabilities add up to 0 .66.

02:55 So we're going to do 1 minus 0 .66, which is 0 .54.

03:02 No, that's wrong.

03:03 0 .34.

03:11 So now that we've done that for a1, let's do it for a2.

03:15 So probability of l2 given a2 is going to be.

03:18 Be we're given that they're late 25 % and 20 % of the time so 0 .25 and 0 .2 and that's 0 .05.

03:31 Conversely we have the probability of no late flights into a2 which is 0 .75 times 0 .8.

03:41 This is 0 .6.

03:44 And then finally we have l1 given a 2.

03:48 There's going to be 1 minus 0 .65.

03:52 Add the other two together and this is 0 .35.