Tree Diagrams Use your keyboard’s arrow keys to move the slides By Henry Mesa Use your keyboard’s arrow keys to move the slides forward (▬►) or backward (◄▬) Use the (Esc) key to exit

Tree Diagrams are another tool that we can use to help us visualize the interaction between probability statements, just like a Venn Diagram. The difference is that the Venn Diagram can not illustrate the probability P(A | B) while the Tree Diagram is made to do just that. Here is an excerpt of what type of probabilities the Venn diagram can illustrate

The Venn diagram represents the “whole” (100%) by creating a rectangle and then it illustrates how that rectangle gets subdivided by the events.

The Tree Diagram represents the “whole” (100%) by creating a node ( a single dot). The tree I am going to draw is the simplest one I can draw based on the previous Venn Diagram.

The Tree Diagram represents the “whole” (100%) by creating a node ( a single dot). Next, you start with one of the events, and either the event occurs or it does not. The lines (branches of the tree) represent either the event occurring or the event does not occur. I write at the end of the branch the name of the event. A Not A Not A

Notice, that a tree has subdivided the entire whole into disjoint parts. I personally call this first part of the tree “stage 1”, where the whole is subdivided into disjoint parts. The next slide gives a different subdivision. A Not A Not A

Now I will continue with my original subdivision (previous slide). Here I subdivided the entire whole into three disjoint parts. I am showing you this slide to make clear what my intent was in the previous slide. A B Now I will continue with my original subdivision (previous slide). C

Next, the subdivisions, get subdivided themselves. A Not B Not A Event A has been subdivided into two parts by introducing event B. So the end labeled B is really…

Next, the subdivisions, get subdivided themselves. A Not B Not A Event A has been subdivided into two parts by introducing event B. So the end labeled B is really… P(B | A)

Warning!!! The B you see is not P(B), but rather the P(B | A): given that we are only looking at the subdivision (sample space) that contains A, find all the items within A that meet event B. B A Not B Not A Now, to complete the picture. I will insert the corresponding probability notation into each branch.

So you can see now that the next stage of the tree, stage 2, is based on the condition of what occurred in stage 1 of the tree. Thus, B can only occur in the upper branch is A occurs. Now B can still occur, if A does not occur. This like saying I pass the first exam or I do not. How does that affect “passing the course.” Any branch beyond stage 1 is always conditional. B P(B | A) A P(A) Not B P(Not B | A) P(B | Not A) B P(Not A) Not A Not B P(Not B | Not A) Notice , I have color coded the Venn diagram. I want to make an association between each region of the Venn and its location on the tree.

Out B A Not B In B Not A Not B P(B | A) P(A) P(Not B | A) P(B | Not A) One way a tree can be read is to start at the node then, follow the paths until you are out. So If I follow the path described below, what is the probability notation that describes the output of that path? The color coding now does correspond to the Venn. Out B P(B | A) P(A AND B) A P(A) Not B P(Not B | A) In P(B | Not A) B P(Not A) Not A Not B P(Not B | Not A) As a formula P(A AND B) = P(A)P(B |A) Now lets finish the rest of the tree output.

Out B A Not B In B Not A Not B P(A)P(B |A) P(B | A) P(A) P(Not B | A) So how can the tree help me answer probability questions? Out B P(B | A) P(A AND B) A P(A) Not B P(A AND Not B) P(Not B | A) In P(B | Not A) B P(Not A AND B) P(Not A) Not A Not B P(Not A AND Not B) P(Not B | Not A) Suppose that I want to answer the question P(A and B)? Think of it as what is the chance I enter into the tree path and come out a the desired result? First you must go down the path that ends at A, a probability of P(A). But afterward you must go through the path labeled B which you have a probability of P(B | A). This means that P(A AND B) is P(A)P(B |A)

An Example

A town has an illness that is expected to affect a population A town has an illness that is expected to affect a population. There is a drug that, if taken before hand may stop the illness. A public announcement is sent out and 45% of the population takes this drug. If you take the drug there is a 79% chance you will not get ill. If you do not take the drug there is a 55% you will not get ill. What percentage of the population will not receive the drug and will get ill? Here is the question using function notation P(NOT drug AND ill)? I want the person to not get the drug and become ill. Notice, the question itself looks like what I asked when you had Venn diagrams, but what is changed is the “tools/probabilities” you have to work with. The first step as always is to write the given probabilities using function notation.

So just like in the Venn situations I create my graph. A town has an illness that is expected to affect a population. There is a drug that, if taken before hand may stop the illness. A public announcement is sent out and 45% of the population takes this drug. If you take the drug there is a 79% chance you will not get ill. If you do not take the drug there is a 55% you will not get ill. What percentage of the population will not receive the drug and will get ill? Here is the question using function notation P(NOT drug AND ill)? I want the person to not get the drug and become ill. The first step as always is to write the given probabilities using function notation. P(drug) = 0.45 P(Not ill | drug) = 0.79 P(Not ill | Not drug) = 0.55 I see that the two other probabilities are conditional and I also notice that I have the probability of the given event by itself, P(drug). To me that signals TREE DIAGRAM. So just like in the Venn situations I create my graph.

A town has an illness that is expected to affect a population A town has an illness that is expected to affect a population. There is a drug that, if taken before hand may stop the illness. A public announcement is sent out and 45% of the population takes this drug. If you take the drug there is a 79% chance you will not get ill. If you do not take the drug there is a 55% you will not get ill. What percentage of the population will not receive the drug and will get ill? Here is the question using function notation P(NOT drug AND ill)? I want the person to not get the drug and become ill. The first step as always is to write the given probabilities using function notation. P(drug) = 0.45 P(Not ill | drug) = 0.79 P(Not ill | Not drug) = 0.55 I will now place the rest of the probabilities along the corresponding branches.

P(drug) = 0.45 P(Not ill | drug) = 0.79 P(Not ill | Not drug) = 0.55 A town has an illness that is expected to affect a population. There is a drug that, if taken before hand may stop the illness. A public announcement is sent out and 45% of the population takes this drug. If you take the drug there is a 79% chance you will not get ill. If you do not take the drug there is a 55% you will not get ill. What percentage of the population will not receive the drug and will get ill? Here is the question using function notation P(NOT drug AND ill)? I want the person to not get the drug and become ill. The first step as always is to write the given probabilities using function notation. P(drug) = 0.45 P(Not ill | drug) = 0.79 P(Not ill | Not drug) = 0.55 So, now I can answer the question. P(Not drug AND ill) = 0.55(0.45) = 0.2475 In Out

P(drug) = 0.45 P(Not ill | drug) = 0.79 P(Not ill | Not drug) = 0.55 A town has an illness that is expected to affect a population. There is a drug that, if taken before hand may stop the illness. A public announcement is sent out and 45% of the population takes this drug. If you take the drug there is a 79% chance you will not get ill. If you do not take the drug there is a 55% you will not get ill. P(drug) = 0.45 P(Not ill | drug) = 0.79 P(Not ill | Not drug) = 0.55 Here is another question: What is percentage of the population will get the drug and get ill? The first step as always is to write the given probabilities using function notation. P(drug AND ill) =? All I have to do know is read the tree to get the appropriate numbers. Out P(drug AND ill) = 0.45(0.21) = 0.0945 In

P(drug) = 0.45 P(Not ill | drug) = 0.79 P(Not ill | Not drug) = 0.55 A town has an illness that is expected to affect a population. There is a drug that, if taken before hand may stop the illness. A public announcement is sent out and 45% of the population takes this drug. If you take the drug there is a 79% chance you will not get ill. If you do not take the drug there is a 55% you will not get ill. P(drug) = 0.45 P(Not ill | drug) = 0.79 P(Not ill | Not drug) = 0.55 Here is another question: What percentage of the population will not take the drug and does not become ill? The first step as always is to write the given probabilities using function notation. P(NOT drug AND NOT ill) =? All I have to do know is read the tree to get the appropriate numbers. P(NOT drug AND NOT ill) = 0.55(0.55) = 0.3025 In Out

P(drug) = 0.45 P(Not ill | drug) = 0.79 P(Not ill | Not drug) = 0.55 A town has an illness that is expected to affect a population. There is a drug that, if taken before hand may stop the illness. A public announcement is sent out and 45% of the population takes this drug. If you take the drug there is a 79% chance you will not get ill. If you do not take the drug there is a 55% you will not get ill. P(drug) = 0.45 P(Not ill | drug) = 0.79 P(Not ill | Not drug) = 0.55 Here is another question: What percentage of the population will become ill? The first step as always is to write the given probabilities using function notation. P(ill) = ? Notice the answer is not 0.21 nor is it 0.45, nor is the sum of the two 0.21 + 0.45. You must truly understand the structure of a tree to answer this question. There are two possible ways to get ill; one involves taking the drug, the other not taking the drug, thus… Out P(ill) = P(ill AND drug OR ill AND NOT drug) Out In = 0.45(0.21) + (0.55)(0.45) = 0.342

Bayes’s Rule

You now see how we can read the tree diagram to answer some questions You now see how we can read the tree diagram to answer some questions. But there is another question that can be answered using the tree diagram, and this type of question can be answered using Bayes’s Rule. I could get very technical here, but this is not the point of this presentation. Consider the direction we have answered the previous questions. We started at the node then worked our way forward toward the ending of the tree. What makes this question different is that, to me, it appears that we are looking at the tree backwards. I will tell you what path I came out of, now you tell me, what path I actually took, or more to the point, give me the probability that my path was….

P(drug) = 0.45 P(Not ill | drug) = 0.79 P(Not ill | Not drug) = 0.55 A town has an illness that is expected to affect a population. There is a drug that, if taken before hand may stop the illness. A public announcement is sent out and 45% of the population takes this drug. If you take the drug there is a 79% chance you will not get ill. If you do not take the drug there is a 55% you will not get ill. P(drug) = 0.45 P(Not ill | drug) = 0.79 P(Not ill | Not drug) = 0.55 Here is another question: A person becomes ill. What is the probability that this person did not take the drug? Notice how this question has already assumed the person is ill; a given. P(Not Drug | ill) = ? I am going to use the formula for conditional probabilities to help answer the question. Out In Out P(Not Drug | ill) = 0.7237

View this presentation as often as you need View this presentation as often as you need. As always don’t be discouraged if this concept does not sit well with you yet. The more exposure you have to this material the easier and more natural this will become.