WEBVTT mathematics/probability/murray
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Hi and welcome back to the probability lectures here on www.educator.com, my name is Will Murray.
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The last time we learned some of the basic terminology for probability and experiments
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about sample spaces and events.
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Today, we are going to keep going with that with the combining events.
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We are going to learn how to build up more complicated events.
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We are going to learn about unions and intersections.
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We are going to learn how to calculate the number of outcomes in events, in intersections, and unions of events.
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We are going to learn some rules of multiplication and addition.
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We are also going to learn about conditional probability and independence.
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There is a lot that is going to be happening in this lesson today.
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We are going to start out with unions of events.
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The union of two events A and B means the set of outcomes in A or B.
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This is one of the most loaded words in the English language, this word or.
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Let us talk about what that means when we say or.
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It means that at least one of A or B is true.
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Or often mean something different in English.
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In English, people some× talk about super salad.
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For example at a restaurant, with your entree you get a super salad but that does not mean you get both,
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that means you get either a soup or a salad.
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That is not what or means in mathematics.
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The super salad is the exclusive or.
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In mathematics, we mean the inclusive or.
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We mean we get one or the other, or you get both.
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Let me draw a little diagram to indicate what I’m talking about when I talk about the inclusive or,
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or the union of events.
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If we have our sample space here, it is the set of all the outcomes, that is our sample space.
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And then we have two events A and B.
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There is A and there is B.
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The union of those two events is all the stuff that is in A or B, or both of them.
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We get all three regions there.
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The stuff that is just A, the stuff that is just in B.
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You also get the stuff that is in A intersect B.
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Everything in there gets counted when we talk about the union.
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When people use the word the or in mathematical settings or in computer science settings,
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they do mean that inclusive or.
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The exclusive or is just not something that is used as often.
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When it is used, people are very explicit to specify it as the exclusive or.
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We definitely, in mathematics, we are talking about the inclusive or.
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When we want to calculate the probability of A union B, we have a formula for it.
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What you do is you calculate the probability of A and then you add the probability of B.
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The problem if you just stop there, is that you have over counted everything that is in both A and B.
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You over counted this region right here in the middle, the intersection of A and B
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because you count that once for A and once for B.
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What you have to do is subtract off that over counting of the intersection.
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You have already counted it twice so you subtracted off so that you have counted it exactly once.
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You subtract off the probability of a intersect b, when you are calculating the probability of the union.
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The easier case there is when they are disjointed events.
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Disjoint means they do not overlap at all.
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Let me draw a little diagram to show disjoint events.
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Here is our sample space and here would be A and here would be B.
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When they do not overlap at all, when there is no intersection between the two events,
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it is much easier to calculate the probability because you just calculate the probability of A and
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then the probability of B and then you add them up because there is no intersection to worry about.
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That is the union of events, let us move on and talk about the intersection of events.
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Again, I will illustrate this with a diagram.
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A intersect B is the set of outcomes that are in both A and B.
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I’m going to setup my little diagram, my sample space here.
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I will draw in two events, A and B.
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That is the whole sample space.
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There is my first event A, there is my second event B.
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I want all the outcomes that are in both A and B.
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That is just this little region in the middle there, that is the intersection.
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A intersect B there.
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It is often helpful when you are calculating things.
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We will see some examples of this, as we get into some of the problems.
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If you have N possible outcomes for one experiment and N possible outcomes for a second independent experiment,
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and you are trying to figure out all the possible outcomes for the combined experiments,
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what you do is you multiply N × N.
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Where N and N possible combined outcomes.
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You will see how that comes in as we do some examples.
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We got a couple more concepts that I want to introduce to you before we get into some examples.
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We have conditional probability and independence.
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Conditional probability, you read this, it is this vertical line notation here, P of A given B is how you read that.
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The conditional probability is written P of A vertical line B but you read that as P of A given B.
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And what it means is, it is the probability that event A is true given that you already know that B is true.
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There is a nice way of calculating it and I can show you this again with the diagram.
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I think it will help you understand where this formula comes from.
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The idea here, let me setup two events A and B.
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There is A and there is B.
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A and B, this is all taking place in a sample space S.
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What we are assuming here is that we already know that B is true.
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We already know that B is true.
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It is restricting our world down to the set of outcomes where B is true.
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We are saying that we are already restricting our self to this red area down here in the sample space.
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What is the probability that A is true given that we are restricting our world to this set of outcomes in B?
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If you are going to be inside B and you want A to be true,
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that means you can only be looking at this region right here because this is the only region inside B that is also in A.
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That region that I just colored in there was exactly A intersect B.
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To find the probability of A given B, we want to find the probability of being in that region,
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the probability of A intersect B.
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The probability of A intersect B but we are restricting herself to a world inside B.
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We divide this by the probability of B.
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We are essentially looking at the probability of the blue region ÷ the probability of the red region.
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That is how we calculate conditional probability.
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We will see some examples, you get some practice with that as we get into some of the problems.
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But there is one more concept I want introduce to quickly introduce to you first which is independence.
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Independence of two events means that suppose you are wondering about whether A is true.
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You are wondering about whether A is going to happen and someone tells you that B is true.
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Someone tells you the B has happened.
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Do you now have any more information about whether A is true?
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If that new information about B being true, that does that make you change your prediction
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about whether A might be true?
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If it does not make you change your prediction then we call those events independent.
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In terms of calculations, this is watching your prediction for A given that B is true.
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That is P of A given B.
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That is if somebody has told you that B is true and you decide to make a prediction about the probability of A.
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This is if you have no information at all.
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You just provide a prediction about P of A with no external information about B.
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If those probabilities would be the same either way, can we say that A and B are independent?
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We say that knowing that B is true does not cause you to change any predictions you might make about A.
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Now, independence is a very dangerous concept.
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It is very misleading and is very frequently mistaken by students.
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Most commonly, students get independence mixed up with being disjointness, what was being disjoint.
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Here is what disjoint means.
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Disjoint means that the probability of A intersect B is 0.
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It means that A and B are completely separate from each other.
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That is not independence.
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Just because they are separate, it does not mean they are independent.
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In fact, it would mean that they are dependent.
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Because if they are disjoint and I tell you that B is true, now you know that A is false.
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If I tell you that B is true, you will get some new information about A.
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Disjointness is not at all the same as independence but people tend to get those mixed up.
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People tend to think that independent means disjoint and disjoint means independence.
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It is not all true.
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We will try to do some examples to highlight the differences between those two.
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To reiterate, if they are disjoint, they are not actually dependent, not independent
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because if you know that one is true then you know that the other one is false.
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Independence means that knowing one is true gives you no new information about the other one.
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Let me show you how this plays out in equations.
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If they are independent, remember independent means P of A is equal to P of A given B.
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P of A given B, here is our formula for P of A given B.
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That is our formula for P of A given B.
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The P of A intersect B/ P of B.
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If they are independent then P of A given B is the same as P of A.
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P of A would be equal to P of A intersect B/ P of B.
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We get this nice formula, if you multiply both sides by P of B, you will get P of A intersect B
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is equal to P of A × P of B.
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That is a really useful multiplication rule except you have to be careful to check that A and B are independent.
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That is the only time you can use this multiplication rule, as if you have independent events.
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Like I said, because students often made mistakes about what independent means,
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you can often get into a lot of trouble by misusing this multiplication rule.
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We will go through some examples to train your intuition to know when an event is independent and when not.
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Let us jump into some examples here.
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First example, we are going to choose a whole number at random, from 1 to 100.
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We want to know, what is the probability that it will be even or greater than 80?
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Let us set up some events here.
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Let me setup a sample space.
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The sample space S is all numbers from 1 to 100, it is a whole number.
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It is 1, 2, 3, up to 100, that is my sample space, all possible outcomes there.
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The events we are interested in are being even and being greater than 80.
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A, I will say is the even numbers.
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That means we have 2, 4, and so on, up to 100 is our last even number there.
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B is that it is greater than 80.
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Everything greater than 80, it says greater than 80 so it is not including 80.
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That is 81 through 100 there, 81, 82, up to 100.
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The key word that we want to focus on here is the or.
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What we want to calculate there is a union, the probability of a union.
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However, what we saw was that to calculate the probability of a union, A union B,
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what we want to do is calculate the probability of A + the probability of B - the probability of A intersect B.
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In order to calculate the union, we need to calculate the intersection.
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We need to go back and calculate the intersection first.
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Let us calculate the sizes of all these sets.
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First of all, how many even numbers are there from 1 to 100?
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There are 50 of these.
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How many numbers are there greater than 80?
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81 through a 100, that is 20 numbers.
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And how many even numbers are there greater than 80?
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That is the intersection, things that are both even and greater than 80.
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That is all the even numbers greater than 80, 82, 84, up to 100.
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There are 10 numbers in there, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100.
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There are 10 of those.
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The size of the union is equal to the size of A.
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I’m really doing this by counting now.
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Let me go back after we work this out by counting and we will do it by probability.
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And you will see a slightly different argument, they will get us to the same answer.
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A + B - the size of the intersection, A intersect B, which we said A was 50, B is 20, the intersection we said is 10.
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That simplifies down to 60, 70 -10 is 60.
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The probability of A union B, that is what we are looking for because it was not or.
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A union B is equal to the size of A union B ÷ the size of the sample space.
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That is 60 ÷ 100 of which reduces down to 3/5.
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That is our probability of getting an even number or a number greater than 80.
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This is a good example of solving a probability problem by counting.
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By counting the sizes of all the sets there are.
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We have counted the size of A, we have counted the size of B.
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We have counted the size of A intersect B.
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We counted the size of A union B and then we divide them all together.
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We could also solve this by probability.
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I like to show you how that goes.
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The arguments are very similar but they will end up at the same answer.
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But sometimes students get mixed up between the two approaches.
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Sometimes you will start out with one approach and then switch over to the other approach,
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or mix the two approaches together and you get confused.
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You end up getting tangled up in getting the wrong answer.
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Let me show you how the two different approaches look and show you what they have in common and how they are different.
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If we do this by probability, if we use this formula up here, the probability of A is the number of even numbers over 100.
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There is 50 even numbers over 100 which is ½.
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The probability of B, B was the event that our pick is greater than 80.
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The number of numbers greater than 80 is 20/ 100 possible choices, that is 1/5.
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The probability of A intersect B is the probability of getting a number that is greater than 80 and is even.
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We said that, where is our A intersect B?
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There it is right there.
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We said that there is 10 numbers that are greater than 80/100 possibilities that is 10/ 100, which is 1/ 10.
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The probability of A union B is the probability of A + the probability of B - the probability of the intersection A intersect B.
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I’m using this formula right here.
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That is ½ + 1/5 – 1/10.
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If I put those all over common denominator which would be 10, I will get 5 + 2 - 1/10 which is 7 -1/6/ 10, which is 3/5.
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Of course, that agrees with the answer we got by counting but it looks a little different in flavor.
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I want to emphasize that a lot of probability problems have a breakdown like this
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where you can solve them strictly by counting, kind of following this technique.
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Or you can solve them by probability, following this technique on the right.
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Each way can work but if you mix up the 2 then you are liable to make mistakes.
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If you kind of mix the two together, it is a recipe for disaster, let me say.
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You want to decide, am I going to do this by counting or am I going to do this by probability?
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Then, stick to your technique and pursue it all the way through.
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Sometimes you can do it by both techniques but if you do, you want to do it completely separately
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and make sure that your answers agree with each other at the end.
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Let me go back and show you quickly the steps we went through here.
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We want to solve it by counting, we first listed our sample space.
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We have listed our two events, the even numbers and the ones greater than 80.
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Those are our two events A and B.
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Count the numbers of outcomes in each one.
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A have 50 outcomes, B had 20 outcomes, and then the intersection had 10 outcomes.
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They are numbers that are both even and greater than 80.
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To get the size of the union, we add up the sizes of A and B and subtract the intersection that turned out to be 60.
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And then the probability, finally we divide 60 ÷ S, that sample space.
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60 ÷ 100 reduced down to 3/5.
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That is solving the whole problem by counting the sizes of the set.
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Instead, if we do it by probability, we calculate the probability of everything as we go along.
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The probability of getting an even number is ½.
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The probability of getting a number greater than 80 is 1/5.
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The probability of getting both even and greater than 80 is 1/10.
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And then we use this generic formula that I gave you in one of the introductory slides, P of A + P of B - P of A intersect B.
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Reduce those fractions down and it reduces down to that same answer 3/5.
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In example 2, we have a combination deal with restaurants where they are offering a combination
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of a main dish and a drink and the dessert.
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We want to figure out how many different types of combination meals we can make.
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The restaurant is offering us 4 different types of drinks, 12 different main dishes, and 3 possible desserts.
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This is an illustration of the multiplication rule.
00:22:33.900 --> 00:22:40.500
The solution here is, how many different types of drinks I can have?
00:22:40.500 --> 00:23:00.900
I can have 4 types of drinks and then independent of that is my choice of main dish.
00:23:00.900 --> 00:23:06.900
For each drink combination, I can have a main dish.
00:23:06.900 --> 00:23:14.900
There are 12 different kinds of main dishes I can have, for each type of drink I might pick.
00:23:14.900 --> 00:23:22.100
For my dessert, I could have 3 different kinds of desserts and I multiply those together.
00:23:22.100 --> 00:23:25.900
This is the multiplication rule, since these are all independent choices.
00:23:25.900 --> 00:23:28.400
What I get for my drink does not affect my main dish.
00:23:28.400 --> 00:23:31.700
What I get for my main dish does not affect my dessert.
00:23:31.700 --> 00:23:48.800
I just multiply those together and get 4 × 3 is 12 × 12 is 144 combinations.
00:23:48.800 --> 00:23:54.000
If you think about how many different possible types of combination meals that restaurants can serve,
00:23:54.000 --> 00:24:01.700
the answer is 3 × 4 × 12 is 144 different kinds of combination meals.
00:24:01.700 --> 00:24:08.300
You can go that restaurant every day, a 144 days in row and order a different combination each day.
00:24:08.300 --> 00:24:14.900
On the 145 day, you have to repeat some combination that you have seen before.
00:24:14.900 --> 00:24:20.200
It is the multiplication rule in action there.
00:24:20.200 --> 00:24:22.100
For example 3, we are going to roll 2 dice.
00:24:22.100 --> 00:24:29.000
And it is useful often when you are rolling dice, it is very useful to think of them as being a red one and a blue one.
00:24:29.000 --> 00:24:37.500
We are being asked what is the probability that the red one shows a 4 and the blue one is odd?
00:24:37.500 --> 00:24:43.600
We are going to set up some events here.
00:24:43.600 --> 00:24:48.300
S is my sample space of all possible outcomes.
00:24:48.300 --> 00:24:52.900
Remember, for rolling two dice, we have this example in the previous lecture.
00:24:52.900 --> 00:24:56.000
You can go back and check it out, if you do not number exactly how it works out.
00:24:56.000 --> 00:24:59.800
But rolling two dice, the red one can be 1 through 6.
00:24:59.800 --> 00:25:03.900
And then, for each one of those, the blue one can be 1 through 6.
00:25:03.900 --> 00:25:09.700
You can get red 1, blue 1.
00:25:09.700 --> 00:25:15.800
You can get red 1, blue 2.
00:25:15.800 --> 00:25:20.500
All the other combinations where the red is 1 and all the combinations where the red is 2,
00:25:20.500 --> 00:25:28.400
all the combinations where red is 3, all the way on up to if the red is 6 and the blue is 6.
00:25:28.400 --> 00:25:35.600
There are 36 possible combinations there, 36 possible outcomes in our sample space.
00:25:35.600 --> 00:25:51.300
Our two events here, our A is the event that the red one is a 4.
00:25:51.300 --> 00:25:54.900
Let us list out those possibilities there, at least the first few.
00:25:54.900 --> 00:26:00.400
The red would be a 4, the blue one could be 1.
00:26:00.400 --> 00:26:14.700
The red one could be a 4, the blue one could be a 2, and so on, up to the red one could be a 4 and the blue one could be a 6.
00:26:14.700 --> 00:26:19.400
Those are all the possibilities there, for the red dice to be a 4.
00:26:19.400 --> 00:26:24.100
B is going to be the event that the blue dice is odd.
00:26:24.100 --> 00:26:31.500
The blue dice is odd.
00:26:31.500 --> 00:26:41.400
Let us think about the number of possibilities there, about how you can have the blue dice being odd.
00:26:41.400 --> 00:26:46.100
You can have a red one being 1 and the blue one is a 1.
00:26:46.100 --> 00:26:57.400
You can have the red one being a 1 and the blue one being not 2 but 3, and so on, up to the last possibility there.
00:26:57.400 --> 00:27:03.400
Be the red one is 6 and the blue one could not be 6, or the last one would be 5, if we list them all out.
00:27:03.400 --> 00:27:06.700
Those are all the possibilities for the blue dice to be odd.
00:27:06.700 --> 00:27:11.200
Remember I said a lot of these probability problems can be solved two different ways.
00:27:11.200 --> 00:27:17.500
You can solve them by counting or you can solve them thinking about probabilities all the way through.
00:27:17.500 --> 00:27:23.500
I would like to try and do this problem both ways so that you can see the distinction.
00:27:23.500 --> 00:27:29.300
And you can get the same answer either way but you can see the different kinds of thinking.
00:27:29.300 --> 00:27:36.200
And in particular, I do not want you to kind of mix up the counting arguments and the probability arguments.
00:27:36.200 --> 00:27:38.700
Let me show you first the counting arguments.
00:27:38.700 --> 00:27:51.100
By counting, we want to find the probability of A intersect B.
00:27:51.100 --> 00:28:01.000
Let us think about how we can count A intersect B and then we will divide it
00:28:01.000 --> 00:28:04.300
by the number of outcomes in our sample space.
00:28:04.300 --> 00:28:10.900
A intersect B means the red dice is 4 and the blue dice is odd.
00:28:10.900 --> 00:28:24.900
The reason I'm using intersection there is because this key word here and, we want A intersect B.
00:28:24.900 --> 00:28:32.100
To be in A intersect B, the red dice has to be 4 and the blue dice has to be odd.
00:28:32.100 --> 00:28:36.000
Red dice can be 4, blue could be 1.
00:28:36.000 --> 00:28:44.800
The red dice could be 4 and the blue could be 3.
00:28:44.800 --> 00:28:53.000
The red dice could be 4 and the blue could be 5.
00:28:53.000 --> 00:28:59.400
Those are the only ways that the red can be 4 and the blue could be odd.
00:28:59.400 --> 00:29:03.200
S is the sample space of all possible outcomes.
00:29:03.200 --> 00:29:13.400
We already figure out that there are 36 of those corresponding to 6 choices for the red dice, 6 choices for the blue dice.
00:29:13.400 --> 00:29:25.500
If we work that out, we got 3/36 which of course simplifies down to 1/12.
00:29:25.500 --> 00:29:31.000
By simply counting things up, we work out that our probability is 1/12.
00:29:31.000 --> 00:29:32.700
That is a counting based argument.
00:29:32.700 --> 00:29:43.100
Let me give you another argument that is based on thinking about probability directly.
00:29:43.100 --> 00:29:47.800
This is a little more subtle but the arithmetic is pretty nice at the end of it.
00:29:47.800 --> 00:29:53.000
It think about these two events, the red dice being a 4 and the blue dice being odd.
00:29:53.000 --> 00:29:59.500
If I roll two dice and I tell you that the red dice is 4, does I tell you anything about the blue dice?
00:29:59.500 --> 00:30:04.200
No, it tells you nothing about the blue dice.
00:30:04.200 --> 00:30:17.400
These events are independent.
00:30:17.400 --> 00:30:26.500
The reason that is true is if I tell you that one of these events happen, if I tell you that the red dice is 4
00:30:26.500 --> 00:30:29.500
and then I say that does change what you predict about the blue dice?
00:30:29.500 --> 00:30:32.500
It does not change what you predict about the blue dice at all.
00:30:32.500 --> 00:30:55.900
Since knowing that one is true does not change the probability of the other.
00:30:55.900 --> 00:30:58.400
Remember, independence does not mean disjoint.
00:30:58.400 --> 00:31:02.300
Disjoint means that the two things cannot happen at the same time.
00:31:02.300 --> 00:31:03.500
This is not what we are talking about.
00:31:03.500 --> 00:31:06.900
We are talking about independence means that if I tell you that one is true,
00:31:06.900 --> 00:31:14.600
it does not change your prediction that the other might be true.
00:31:14.600 --> 00:31:21.200
Because these two events are independent, we have an easy way to calculate the probability of the intersection.
00:31:21.200 --> 00:31:22.500
This is from that opening slide.
00:31:22.500 --> 00:31:28.000
If you do not remember this, just go back a few slides and you will see it.
00:31:28.000 --> 00:31:39.800
The probability of A intersect B, when they are independent is the probability of A × the probability of B.
00:31:39.800 --> 00:31:43.100
That is really using independence very heavily right there.
00:31:43.100 --> 00:31:46.700
This equation would not be true if they were not independent.
00:31:46.700 --> 00:31:50.800
But let us think about the probability of A and B independently.
00:31:50.800 --> 00:31:58.800
The probability that the red dice is 4, if you just forget that you are rolling a blue dice,
00:31:58.800 --> 00:32:04.300
the probability that you roll on dice and you get a 4 is 1 out of 6
00:32:04.300 --> 00:32:07.900
because there are 6 possible outcomes that one of them gives you a 4.
00:32:07.900 --> 00:32:10.700
What is the probability that the blue dice is odd?
00:32:10.700 --> 00:32:16.700
Again, for get that you are rolling a red dice, the probability that the blue dice is odd,
00:32:16.700 --> 00:32:18.900
there are 6 possible outcomes.
00:32:18.900 --> 00:32:26.100
3 of them will give you an odd number, 3 out of 6 is ½.
00:32:26.100 --> 00:32:39.000
We can fill those in here, P of A × P of B is 1/6 × ½, multiply those together and you get 112.
00:32:39.000 --> 00:32:42.900
That is the probability that both of those events will occur.
00:32:42.900 --> 00:32:46.200
That is the probability of the intersection of those events.
00:32:46.200 --> 00:32:50.600
And notice, that agrees with our earlier answer of 112.
00:32:50.600 --> 00:32:52.500
We got 112 either way.
00:32:52.500 --> 00:32:59.100
But using two quite different arguments, one is based on counting and one is based on the theory of independence.
00:32:59.100 --> 00:33:02.700
Let me recap what we did there.
00:33:02.700 --> 00:33:07.100
We first listed our sample space which is all the possible combinations of the two dice.
00:33:07.100 --> 00:33:13.400
There are 36 combinations depending on what the red shows and what the blue shows.
00:33:13.400 --> 00:33:18.700
A is the probability, A is the event that the red dice show a 4.
00:33:18.700 --> 00:33:22.000
We have listed out the 6 possible combinations there.
00:33:22.000 --> 00:33:28.600
We also calculated the probability of A is 1/6 because if you forget that you are rolling a blue dice,
00:33:28.600 --> 00:33:32.400
the probability that the red dice will be 4 will be 1 out of 6.
00:33:32.400 --> 00:33:40.200
The probability that the blue dice is odd, if you list out all those possible combinations,
00:33:40.200 --> 00:33:42.500
I do not think I ever bother to count those.
00:33:42.500 --> 00:33:49.700
There will be 18 combinations but we found the probability just to be ½
00:33:49.700 --> 00:33:55.700
because if you forget that you are rolling a red dice then the probability that the blue dice will be odd,
00:33:55.700 --> 00:33:57.300
there are 3 ways for it to be odd.
00:33:57.300 --> 00:34:01.200
1, 3, and 5 ÷ 6 possible combinations.
00:34:01.200 --> 00:34:05.300
3/6 or about ½ came from.
00:34:05.300 --> 00:34:13.900
If we do counting argument here, we just try to count the size of A intersect B.
00:34:13.900 --> 00:34:19.100
Here the outcomes in A intersect B meaning the red dice is 4 and the blue dice is odd.
00:34:19.100 --> 00:34:22.200
There are three ways for that to happen, 4-1, 4-3, and 4-5.
00:34:22.200 --> 00:34:29.800
Out of 36 total possible outcomes, we get t3 out of 36 is 112.
00:34:29.800 --> 00:34:34.200
If we use a probability theory though, if we use the theory of independence,
00:34:34.200 --> 00:34:38.900
knowing that the red dice is 4 does not tell you anything about the blue dice.
00:34:38.900 --> 00:34:44.400
And knowing that the blue dice is odd, does not tell you anything about the red dice.
00:34:44.400 --> 00:34:46.600
That means that these are independent.
00:34:46.600 --> 00:34:52.400
And we have another way to calculate probabilities of intersections when the events are independent.
00:34:52.400 --> 00:34:55.200
You can just multiply their probabilities.
00:34:55.200 --> 00:35:01.000
We already calculated the probabilities to be 1/6 to ½, multiply those together and get 112
00:35:01.000 --> 00:35:10.000
which agrees with the answer we got using the other argument.
00:35:10.000 --> 00:35:15.400
In example 4, we are going to flip a coin 4 × and we want to find the probability that
00:35:15.400 --> 00:35:20.000
there would be at least 3 heads given that the first two flips are heads.
00:35:20.000 --> 00:35:25.300
This is an example where we are given some partial information about a problem and
00:35:25.300 --> 00:35:34.600
we want to calculate the probability of an event based on knowing that another event is true,
00:35:34.600 --> 00:35:39.600
based on that partial information that we have.
00:35:39.600 --> 00:35:43.700
Let me set up some events here, our sample space.
00:35:43.700 --> 00:35:49.800
S is all the possible things that can happen when you flip a coin 4 ×.
00:35:49.800 --> 00:35:55.000
When you think about it, each time you flip there is two possible things that can happen and
00:35:55.000 --> 00:36:01.900
if you string those together, it is 2 × 2 × 2 × 2.
00:36:01.900 --> 00:36:12.100
You can get head-head-head-head, head-head-head-tail, and so on.
00:36:12.100 --> 00:36:18.400
There is two choices for each one of those spot, each one of those 1st, 2nd, 3rd, and 4th spots.
00:36:18.400 --> 00:36:24.700
There are 16 possible outcomes in our sample space here.
00:36:24.700 --> 00:36:37.100
We want to identify the events that the problem is asking us about, 16 possible outcomes.
00:36:37.100 --> 00:36:40.700
I see this we are not given that means we are going to use conditional probability.
00:36:40.700 --> 00:36:42.600
Let me set up my events here.
00:36:42.600 --> 00:36:49.800
A is the probability that there will be at least 3 heads.
00:36:49.800 --> 00:36:55.000
When we try to list those out, at least 3 heads.
00:36:55.000 --> 00:36:59.700
How can we get at least 3 heads?
00:36:59.700 --> 00:37:04.400
List out those possibilities we can get head-head-head-tail.
00:37:04.400 --> 00:37:07.800
We can get head-head-tail-head.
00:37:07.800 --> 00:37:09.800
That would have at least 3 heads.
00:37:09.800 --> 00:37:18.800
Head-tail-head-head, tail-head-head-head, or we could get all 4 heads.
00:37:18.800 --> 00:37:22.700
That would still count as at least 3 heads.
00:37:22.700 --> 00:37:25.500
Those are all the possible ways to get at least 3 heads.
00:37:25.500 --> 00:37:30.100
B is the first two flips are head.
00:37:30.100 --> 00:37:34.600
That was our other condition given in the problem.
00:37:34.600 --> 00:37:43.100
The first two flips are heads.
00:37:43.100 --> 00:37:46.700
When we list all the strings that would give us the first 2 flips are heads.
00:37:46.700 --> 00:38:00.700
Head-head-head-head, head-head-head-tail, head-head-tail-head, head-head-tail-tail.
00:38:00.700 --> 00:38:03.500
Those are all the strings for the first two flips being heads.
00:38:03.500 --> 00:38:09.600
And now we are interested in finding the probability of A given B.
00:38:09.600 --> 00:38:12.600
Let me remind you of the formula for that.
00:38:12.600 --> 00:38:17.500
Probability of A given B is a conditional probability problem.
00:38:17.500 --> 00:38:25.500
Probability of A given B is a probability of A intersect B ÷ the probability of B.
00:38:25.500 --> 00:38:28.000
That was the formula we had in one of the introductory slide.
00:38:28.000 --> 00:38:32.400
Just flip back a few slides and you will see that.
00:38:32.400 --> 00:38:37.400
That means I have to figure out what the probability of A intersect B is.
00:38:37.400 --> 00:38:40.400
I need to find A intersect B.
00:38:40.400 --> 00:38:47.000
I just scan over those list A intersect B of A and B and see which ones are in both.
00:38:47.000 --> 00:38:52.500
I see a head-head-head-tail, HHHT.
00:38:52.500 --> 00:38:54.700
I lost one of my heads there.
00:38:54.700 --> 00:39:06.800
HHHT, I see HHTH, HTHH does not work, THHH does not work.
00:39:06.800 --> 00:39:12.600
HHHH also fixed in both of them.
00:39:12.600 --> 00:39:18.600
I see three of those outcomes are in A intersect B.
00:39:18.600 --> 00:39:24.200
The probability of A intersect B is 3 out of 16 possible outcomes.
00:39:24.200 --> 00:39:34.600
The probability of B is, let us see, there are 1, 2, 3, 4, out of 16 possible outcomes.
00:39:34.600 --> 00:39:47.000
4 out of 16 I can multiply top and bottom by 16 there to simplify things and get ¾.
00:39:47.000 --> 00:39:50.600
Let me show you again where everything in there came from.
00:39:50.600 --> 00:39:53.900
We get this probability problem - flip the coin 4 times.
00:39:53.900 --> 00:39:56.000
Right away, I got a sample space here.
00:39:56.000 --> 00:40:02.200
I got 16 possible outcomes and describing all the possible strings of heads and tails that we can get.
00:40:02.200 --> 00:40:05.900
What is the probability that there will be at least 3 heads?
00:40:05.900 --> 00:40:10.000
I’m going to list that out as one of my events, at least 3 heads.
00:40:10.000 --> 00:40:12.800
They are all the ways that you can get at least 3 heads.
00:40:12.800 --> 00:40:16.100
It looks like there is 5 of them.
00:40:16.100 --> 00:40:24.100
Given that the first two flips are heads, they are all the ways to get the first two flips being heads.
00:40:24.100 --> 00:40:26.800
Listed all those out and I got 4 of them.
00:40:26.800 --> 00:40:33.300
The key word here is given, that tells me that I'm going to be calculating conditional probability.
00:40:33.300 --> 00:40:35.100
That is what this one is all about, it is conditional probability.
00:40:35.100 --> 00:40:37.600
I wrote down my conditional probability formula.
00:40:37.600 --> 00:40:42.800
P of A given B is P of A intersect B/ P of B.
00:40:42.800 --> 00:40:48.800
In order to calculate that, I know what P of B is because of I have already looked at B.
00:40:48.800 --> 00:40:51.900
I have to find A intersect B.
00:40:51.900 --> 00:40:57.900
I looked at A and B, and I try to figure out which of the strings are in both.
00:40:57.900 --> 00:41:00.200
I found these 3 strings in both of them.
00:41:00.200 --> 00:41:04.000
Which means the probability of A intersect B is 3/16.
00:41:04.000 --> 00:41:09.200
That 16 comes from here, the t3 comes from here.
00:41:09.200 --> 00:41:12.300
And the probability of B is 4/16.
00:41:12.300 --> 00:41:14.500
The 4 comes from here.
00:41:14.500 --> 00:41:23.500
When you simplify that down, you get the answer is ¾.
00:41:23.500 --> 00:41:28.000
For example 5, we are going to roll two dice, we would think of them as red and blue.
00:41:28.000 --> 00:41:38.400
We are going to define the events that the red dice is 3, that is event A, B is that the total of 7 and C is that the total is 8.
00:41:38.400 --> 00:41:42.500
There is a whole bunch of possible combinations here.
00:41:42.500 --> 00:41:48.900
And the problem is asking us to calculate the conditional probabilities of all these combinations.
00:41:48.900 --> 00:41:54.700
B given A, A given B, C given A, A given C, and so on.
00:41:54.700 --> 00:41:57.700
We need to calculate a lot of probabilities here.
00:41:57.700 --> 00:42:02.100
Remember our formula for conditional probability, let me just remind you what is.
00:42:02.100 --> 00:42:13.400
P of A given B is the probability of A intersect B ÷ the probability B.
00:42:13.400 --> 00:42:17.800
We are being asked to find the conditional probabilities of everything in sight.
00:42:17.800 --> 00:42:22.500
Which means we are going to need the probabilities of all the basic events.
00:42:22.500 --> 00:42:25.200
We will also need the probabilities and of all their intersections.
00:42:25.200 --> 00:42:29.400
Let us go ahead and calculate all these out.
00:42:29.400 --> 00:42:33.500
The red dice is 3.
00:42:33.500 --> 00:42:35.700
The probability of A, we are going to start with that.
00:42:35.700 --> 00:42:39.900
The red dice is 3, that is a probability of 1/6.
00:42:39.900 --> 00:42:44.300
One way to think about that is there are 6 possible outcomes where the red dice is 3
00:42:44.300 --> 00:42:50.900
because the blue dice can be any of 6 different possibilities ÷ 36 possible outcomes in total.
00:42:50.900 --> 00:42:53.600
That is not the most efficient way to think about it.
00:42:53.600 --> 00:42:58.300
The most efficient way to think about it is to just forget that you are rolling a blue dice and
00:42:58.300 --> 00:43:00.500
say what is the probability that the red dice is 3?
00:43:00.500 --> 00:43:03.200
It is 1/6, the total is 7.
00:43:03.200 --> 00:43:07.100
We saw this one in an earlier example in the previous lecture.
00:43:07.100 --> 00:43:12.000
Check back in that example, if you do not remember it because I will do it quickly now.
00:43:12.000 --> 00:43:13.100
The total is 7.
00:43:13.100 --> 00:43:16.200
There are 6 ways to get the total being 7.
00:43:16.200 --> 00:43:29.700
You can get a 1-6, you can get a 2-5, you can get a 3-4, 4-3, 5-2, or 6-1.
00:43:29.700 --> 00:43:32.100
There are 6 ways you can get a total of 7.
00:43:32.100 --> 00:43:44.800
There is 36 possible rolls in all, the probability of B is 6/36 but that just simplifies down to 1/6.
00:43:44.800 --> 00:43:46.700
C is that the total is 8.
00:43:46.700 --> 00:43:50.300
Let me write down quickly the combinations that give you a total of 8.
00:43:50.300 --> 00:43:55.800
You can get anything if there is a 1 on the first dice.
00:43:55.800 --> 00:44:07.600
You have to get 2-6, or 3-5, or 4-4, or 5-3, or 6-2.
00:44:07.600 --> 00:44:13.400
Those are all the possibilities and there is 5 of those.
00:44:13.400 --> 00:44:21.400
Out of 36 possibilities, possible outcomes and all, the probability of C is 5/36.
00:44:21.400 --> 00:44:29.900
We are also going to have to find the probabilities of all the intersections of these combinations of events.
00:44:29.900 --> 00:44:36.000
Let us go ahead and calculate all those and then we can start to work on the conditional probabilities.
00:44:36.000 --> 00:44:49.500
P of A intersect B means the red dice is 3 and the total is 7, that is the probability.
00:44:49.500 --> 00:44:54.700
The only way to get the red dice in being 3 and the total being 7 is,
00:44:54.700 --> 00:45:00.000
if the red dice is 3 and the blue dice would have to be 4 for that to work.
00:45:00.000 --> 00:45:05.500
The only way to get that, there is only one possible outcome which would be the red dice is 3, blue dice is 4.
00:45:05.500 --> 00:45:10.200
One possible outcome out of 36 things that can happen.
00:45:10.200 --> 00:45:14.000
That is 1 out of 36.
00:45:14.000 --> 00:45:21.500
Let us find the probability of A intersect C, that is kind of similar.
00:45:21.500 --> 00:45:25.900
That means the red dice is 3 and the total is 8.
00:45:25.900 --> 00:45:34.400
The only way you can get that would be if the red dice were 3 and the blue dice were 5.
00:45:34.400 --> 00:45:38.500
Again, that is one possible outcome out of 36 in total.
00:45:38.500 --> 00:45:44.300
1 out of 36 chance that both of those events will be true.
00:45:44.300 --> 00:45:48.400
Finally, the probability of B intersect C.
00:45:48.400 --> 00:45:53.400
We need this to calculate these conditional probabilities.
00:45:53.400 --> 00:45:59.200
Think about what that means, B intersect C means B and C are both true.
00:45:59.200 --> 00:46:01.700
B means the total is 7.
00:46:01.700 --> 00:46:03.600
C means the total is 8.
00:46:03.600 --> 00:46:12.400
The probability that the total is 7 and the total is 8, those cannot both occur at the same time.
00:46:12.400 --> 00:46:15.100
That probability is 0.
00:46:15.100 --> 00:46:19.500
There is no outcome that gives you the total being 7 and the total being 8.
00:46:19.500 --> 00:46:22.600
That probability is just 0.
00:46:22.600 --> 00:46:25.600
Finally, we were able to calculate our conditional probabilities.
00:46:25.600 --> 00:46:31.900
We are going to use this formula over and over again here.
00:46:31.900 --> 00:46:41.000
The probability of B given A is the probability of A intersect B or B intersect A, those are the same thing.
00:46:41.000 --> 00:46:46.800
A intersect B over the probability of A.
00:46:46.800 --> 00:46:54.800
I have calculated all these things up above so that is 1/36 ÷ the probability of A.
00:46:54.800 --> 00:47:05.200
Over here we got 1/6, that flips up to 6/36 which is 1/6.
00:47:05.200 --> 00:47:09.100
That is the probability of B given A.
00:47:09.100 --> 00:47:15.700
Conditional probability of A given B is the probability of A intersect B again.
00:47:15.700 --> 00:47:19.200
But now we divide by the probability of B.
00:47:19.200 --> 00:47:27.200
But we get the same numbers, 1/36 ÷ probability of B was also 1/6.
00:47:27.200 --> 00:47:32.200
Again, it simplifies down to 1/6.
00:47:32.200 --> 00:47:37.400
The probability of C given A is, I’m following my formula here.
00:47:37.400 --> 00:47:45.400
The probability of C intersect A which is the same as A intersects C ÷ the probability of A.
00:47:45.400 --> 00:47:50.300
A intersects C, here it is up here 1/36.
00:47:50.300 --> 00:47:53.900
The probability of A is 1/6, we get the same numbers again.
00:47:53.900 --> 00:47:59.600
It simplifies down to the same answer 1/6.
00:47:59.600 --> 00:48:04.100
Let us write that 6 a little bigger so that people can read it.
00:48:04.100 --> 00:48:15.600
The probability of A given C, that is the probability of A intersect C ÷ the probability of C, this time.
00:48:15.600 --> 00:48:23.300
1/36 ÷ the probability of C here is 5/36.
00:48:23.300 --> 00:48:34.900
5/36, if we multiply top and bottom by 36 I will get 1/5.
00:48:34.900 --> 00:48:42.800
That is kind of our first interesting surprise there because the probability of C given A and
00:48:42.800 --> 00:48:48.400
the probability of A given C turned out not to be equal to each other.
00:48:48.400 --> 00:48:51.400
We have to be careful from now on.
00:48:51.400 --> 00:48:54.400
Which one we are calculating because they might not be equal to each other.
00:48:54.400 --> 00:49:01.000
B given A and A given B did turn out to be equal but apparently that was just a fluke for those two events.
00:49:01.000 --> 00:49:09.500
In general, the conditional probability does matter which order you write those events in.
00:49:09.500 --> 00:49:20.000
The probability of C given B, that is the probability of B intersect C ÷ the probability of B.
00:49:20.000 --> 00:49:25.200
The probability of B intersect C, we discover was 0.
00:49:25.200 --> 00:49:32.400
0 ÷ 1/6 is still just going to be 0.
00:49:32.400 --> 00:49:35.400
There is no probability of C given B.
00:49:35.400 --> 00:49:42.200
Of course that makes sense because if I tell you that the total is 7 and then I ask what the probability is
00:49:42.200 --> 00:49:46.700
if the total is 8, then of course you should say 0.
00:49:46.700 --> 00:49:50.200
Because if you already know the total is 7, it cannot be 8.
00:49:50.200 --> 00:49:53.600
Let us check the probability of B given C.
00:49:53.600 --> 00:50:06.200
It is again, the probability of B intersect C ÷ the probability of C which again is 0 ÷ 5/36.
00:50:06.200 --> 00:50:13.900
But that numerator does not really matter because we get 0 anyway.
00:50:13.900 --> 00:50:17.000
First, it looked like the answers will all going to 1/6 and then we got into
00:50:17.000 --> 00:50:20.300
some more interesting numbers and they started to change.
00:50:20.300 --> 00:50:23.300
Let us just recap where those all came from.
00:50:23.300 --> 00:50:24.400
We are rolling two dice.
00:50:24.400 --> 00:50:31.600
We know that if we roll a two dice, there are 36 possible outcomes in our experiment because the red
00:50:31.600 --> 00:50:35.400
and blue dice there could each be 6 possible numbers.
00:50:35.400 --> 00:50:41.700
We are given three events and we have to calculate all these conditional probabilities.
00:50:41.700 --> 00:50:48.400
A good thing to do at first was to calculate the probability of each of these events solo, which we did.
00:50:48.400 --> 00:50:54.700
We found that there is 1/6 chance that the red dice is 3, 1/6 chance that the total is 7.
00:50:54.700 --> 00:51:00.200
We got that by adding up all the total ways that we can get 7.
00:51:00.200 --> 00:51:03.900
5 out of 36 chance that the total is 8.
00:51:03.900 --> 00:51:10.500
We get that by adding up the 5 possible ways that you can get a total of 8 there.
00:51:10.500 --> 00:51:17.400
We got the probabilities of all the individual events and we want to find the probabilities of all the combinations.
00:51:17.400 --> 00:51:23.100
The reason that I let those is because I was looking ahead to this conditional probability formula.
00:51:23.100 --> 00:51:27.800
I see that it requires calculating the probability of intersections.
00:51:27.800 --> 00:51:33.300
That is why I went ahead and calculated all these intersections ahead of time, A intersect B.
00:51:33.300 --> 00:51:38.800
The only way to get a red dice 3 and a total of 7, as if the blue dice is 4.
00:51:38.800 --> 00:51:41.000
There is only one outcome that gives you that.
00:51:41.000 --> 00:51:44.100
That is why we get 1 out of 36 there.
00:51:44.100 --> 00:51:52.000
Red dice is 3 and the total is 8 would have to be a 3 and a 5, 1 out of 36 again.
00:51:52.000 --> 00:51:57.000
The total is 7 and the total is 8, there is no way those can both be true.
00:51:57.000 --> 00:51:59.200
There is a 0 probability there.
00:51:59.200 --> 00:52:01.500
We need to calculate the conditional probabilities.
00:52:01.500 --> 00:52:09.700
What I was doing is just invoking this formula up here over and over again.
00:52:09.700 --> 00:52:14.100
Just substituting in and out the different combinations of a's, b's, and c's.
00:52:14.100 --> 00:52:17.100
I just substituted in all the combinations of a's, b's, and c's.
00:52:17.100 --> 00:52:24.300
And then, I plug in all the probabilities of the intersections that I calculated above and the denominators.
00:52:24.300 --> 00:52:31.200
I have plugged in the probabilities of the raw events and simplify them down into all these fractions.
00:52:31.200 --> 00:52:37.000
The interesting thing that happened here was that the probability of C given A and
00:52:37.000 --> 00:52:39.900
the probability of A given C turned out not to be equal.
00:52:39.900 --> 00:52:43.500
You do have to be very careful about the order of these things.
00:52:43.500 --> 00:52:50.100
Also, the probability of either B or C given the other one, turned out to be 0.
00:52:50.100 --> 00:52:53.700
And that of course is confirmed by the calculation.
00:52:53.700 --> 00:53:00.600
It also confirms with your intuition because if I tell you that the total is 7, If I say that the total is 7,
00:53:00.600 --> 00:53:03.400
there is really no way that the total can 8.
00:53:03.400 --> 00:53:11.900
As soon as I tell you that the total is 7, then you say the probability that it is 8 is 0.
00:53:11.900 --> 00:53:18.500
I want you to hang onto all these probabilities because we are going use them again for the next example.
00:53:18.500 --> 00:53:24.900
Example 6 is using this very same experiment and we are going to ask whether
00:53:24.900 --> 00:53:28.700
all these events are independent of each other.
00:53:28.700 --> 00:53:36.200
Instead of recalculating these probabilities, I’m just going to refer to all my answers from this example.
00:53:36.200 --> 00:53:44.400
So hang on to these, make sure you understand them, and then will move on to the next example.
00:53:44.400 --> 00:53:48.800
In example 6 here, this is really referring back to example 5.
00:53:48.800 --> 00:53:51.000
It is the same experiment and the same events.
00:53:51.000 --> 00:53:54.300
We are going to roll two dice, red and blue.
00:53:54.300 --> 00:53:58.700
We are going to define the events A is the event that the red dice is 3.
00:53:58.700 --> 00:54:00.700
B is the event that the total is 7.
00:54:00.700 --> 00:54:03.000
C is the event that the total is 8.
00:54:03.000 --> 00:54:08.300
We saw those before, we calculated the probabilities of each one.
00:54:08.300 --> 00:54:11.900
In fact, let me go ahead and remind you what they were.
00:54:11.900 --> 00:54:16.800
If you do not know how we got them, if you have not just looked at example 5,
00:54:16.800 --> 00:54:21.200
go back and check through example 5 and you will see where all these come from.
00:54:21.200 --> 00:54:36.100
The probability of A was 1/6, the probability of B was also 1/6, and the probability of C is 5 out of 36.
00:54:36.100 --> 00:54:41.900
And then we also calculated several conditional probabilities that are going to be useful here
00:54:41.900 --> 00:54:45.200
when we are determining independence.
00:54:45.200 --> 00:54:48.200
I will just write down some of the useful ones here.
00:54:48.200 --> 00:54:55.100
The probability of A given B, we calculated this last time, in the last example, in example 5.
00:54:55.100 --> 00:54:58.400
It turn out to be 1/6.
00:54:58.400 --> 00:55:04.700
The probability of A given C, I'm skipping ahead to the ones we are going to be using.
00:55:04.700 --> 00:55:09.900
The probability of A given C turned out to be 1/5.
00:55:09.900 --> 00:55:19.000
The probability of B given C, because those were disjoint events, it turned out to be 0
00:55:19.000 --> 00:55:21.800
because they cannot both happen at the same time.
00:55:21.800 --> 00:55:25.400
We are being asked, are events A and B independent?
00:55:25.400 --> 00:55:28.400
What about A and C and what about B and C?
00:55:28.400 --> 00:55:43.800
For A and B, remember independence by definition means the probability of A is equal to the probability of A given B.
00:55:43.800 --> 00:55:50.700
Intuitively, it means that if you calculate the probability of A with information or
00:55:50.700 --> 00:55:57.500
if you calculate the probability of A with the information that B is true, you get the same answers either way.
00:55:57.500 --> 00:55:58.600
Let us just check those out.
00:55:58.600 --> 00:56:02.600
The probability of A, I wrote down was 1/6.
00:56:02.600 --> 00:56:04.700
The probability of A given G is 1/6.
00:56:04.700 --> 00:56:27.100
Those are equal so that tells us that A and B are independent.
00:56:27.100 --> 00:56:28.800
That is our answer for A and B.
00:56:28.800 --> 00:56:31.800
Let us look at some of the others.
00:56:31.800 --> 00:56:38.100
What about A and C, we are going to look at A and C.
00:56:38.100 --> 00:56:49.400
Let us calculate the probability of A and we will check whether it is equal to the probability of A given C.
00:56:49.400 --> 00:56:56.300
The probability of A is 1/6, we got that up above.
00:56:56.300 --> 00:57:01.000
Is that equal to the probability of A given C 1/5?
00:57:01.000 --> 00:57:02.100
No, that is not equal.
00:57:02.100 --> 00:57:05.100
1/6 is not equal to 1/5.
00:57:05.100 --> 00:57:13.700
That tells us that A and C are not independent.
00:57:13.700 --> 00:57:18.900
They are dependent.
00:57:18.900 --> 00:57:22.700
I will try to give you some intuitive feel for that in a moment.
00:57:22.700 --> 00:57:24.700
I want to go through and do all the calculations first.
00:57:24.700 --> 00:57:32.700
Then, we will come back and see if we can make some kind of intuitive interpretation of this.
00:57:32.700 --> 00:57:38.400
That was for A and C, now B and C.
00:57:38.400 --> 00:57:43.400
It is tempting to say they are independent because if you are thinking like some students think,
00:57:43.400 --> 00:57:48.100
they say that total is 7 and total is 8, those cannot both happen.
00:57:48.100 --> 00:57:53.400
You might think all that means independent, that does not mean independent but that means disjoint.
00:57:53.400 --> 00:57:56.100
Let us actually calculate it using the probabilities.
00:57:56.100 --> 00:58:02.400
Is the probability of B equal to the probability of B given C?
00:58:02.400 --> 00:58:04.600
That is our formal definition of independence.
00:58:04.600 --> 00:58:06.600
Let us check and see if those are equal.
00:58:06.600 --> 00:58:09.600
The probability of B is 1/6.
00:58:09.600 --> 00:58:13.200
Is that equal to the probability of B given C is 0.
00:58:13.200 --> 00:58:14.000
Is 1/6 = 0?
00:58:14.000 --> 00:58:17.600
No, it is not equal to 0.
00:58:17.600 --> 00:58:29.400
That tells us that B and C are not independent, they are dependent.
00:58:29.400 --> 00:58:32.200
We did that all using calculations.
00:58:32.200 --> 00:58:35.200
Using the calculations that we got from example 5.
00:58:35.200 --> 00:58:39.200
If you do not know where these numbers came from, just go back and watch example 5.
00:58:39.200 --> 00:58:47.200
The one just before this one and you will see where I got all these numbers, the 1/6, 1/6, and 5 out of 36 for A, B, and C.
00:58:47.200 --> 00:58:52.200
We calculated the conditional probabilities 1/6, 1/5, and 0.
00:58:52.200 --> 00:58:57.400
We calculated some others in example 5 but I just extracted the ones I would need here.
00:58:57.400 --> 00:59:03.500
To check independence using calculations, you just check whether the conditional probability
00:59:03.500 --> 00:59:10.900
is equal to the probability without making the assumption that the other event is true.
00:59:10.900 --> 00:59:16.700
That is what I check, I just check the numbers each time and if they were equal, I said they are independent.
00:59:16.700 --> 00:59:22.700
If they are not equal, I said they are dependent.
00:59:22.700 --> 00:59:30.700
That kind of answers the problem but it would be nice to have some intuitive reason for why these might be true.
00:59:30.700 --> 00:59:32.600
For A and B, it is not so obvious.
00:59:32.600 --> 00:59:40.600
If the red dice is 3, what is that tell you about the total being 7?
00:59:40.600 --> 00:59:49.200
And the answer is it does not really tell you anything because the total being 7 was a 1/6 chance anyway.
00:59:49.200 --> 00:59:56.900
If the red dice is 3, then I know to get the total being 7, I need the blue dice to be a 4 and there is a 1/6 chance of that.
00:59:56.900 --> 01:00:03.200
It is telling me that the red dice is 3 gives me no new information about whether the total was 7.
01:00:03.200 --> 01:00:08.300
That is why A and B are independent.
01:00:08.300 --> 01:00:09.000
How about A and C?
01:00:09.000 --> 01:00:11.700
The red dice is 3 and the total is 8?
01:00:11.700 --> 01:00:17.900
You might think that that is similar to the previous one, to the red dice is 3 and the total is 7, but it is not.
01:00:17.900 --> 01:00:20.100
Here is why.
01:00:20.100 --> 01:00:26.700
To get the total being 8, let us think about how that can happen.
01:00:26.700 --> 01:00:29.200
How can the total be 8?
01:00:29.200 --> 01:00:36.100
You can get 2-6, you can get a 3-5, you can get a 4-4.
01:00:36.100 --> 01:00:37.700
These are always to get a total being 8.
01:00:37.700 --> 01:00:44.100
You can get a 5-3, or you can get a 6-2.
01:00:44.100 --> 01:00:50.900
In particular, for the total to be 8, you cannot roll a 1 on either dice.
01:00:50.900 --> 01:01:06.900
If I tell you that A is true, if I tell you that the red dice is 1, what I really told you is in particular that the red dice is not a 1.
01:01:06.900 --> 01:01:13.800
It is slightly more likely for the total to be 8 because we know now that the red dice is not a 1.
01:01:13.800 --> 01:01:23.700
That is why P of A given C is slightly higher than P of A alone.
01:01:23.700 --> 01:01:29.200
I guess actually that means I have told you that the total is 8,
01:01:29.200 --> 01:01:37.700
that means that the red dice cannot be 1 which makes it slightly more likely for it to be 3.
01:01:37.700 --> 01:01:41.800
That is why we get a 1/5 over here and a 1/6 over there.
01:01:41.800 --> 01:01:45.400
I’m giving you a little bit of information when I tell you that the total is 8.
01:01:45.400 --> 01:01:52.000
I’m saying, I’m really telling you that the red dice cannot be 1 which makes it slightly more likely for it to be 3.
01:01:52.000 --> 01:01:56.400
B and C is easier to understand intuitively.
01:01:56.400 --> 01:02:02.700
If I tell you that the total is 8, I'm giving you some information about whether the total is 7.
01:02:02.700 --> 01:02:04.900
I told you that the total is not 7.
01:02:04.900 --> 01:02:07.400
I changed the probability on you.
01:02:07.400 --> 01:02:12.600
As soon as I tell you that the total is 8, you know for sure that the probability is not 7.
01:02:12.600 --> 01:02:14.500
I have really changed your thinking.
01:02:14.500 --> 01:02:18.900
Those are really dependent events, they do affect each other.
01:02:18.900 --> 01:02:23.400
That is kind of an intuitive reason why we got these answers that we did.
01:02:23.400 --> 01:02:30.500
Why A and C are dependent, B and C are dependent, but A and B are independent.
01:02:30.500 --> 01:02:34.600
There is a lot of twist and turns into this probability and independence.
01:02:34.600 --> 01:02:36.800
I hope you had fun playing with those.
01:02:36.800 --> 01:02:39.300
We are going to move on to the next lesson.
01:02:39.300 --> 01:02:43.400
I want to say thank you for joining us today for the probability lectures on www.educator.com.
01:02:43.400 --> 01:02:47.000
My name is Will Murray, thanks for joining, bye.