WEBVTT mathematics/math-analysis/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about probability.
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Consider if we wanted to know the chance of rain today, the odds of winning a bet, or how likely a medicine is to cure a disease.
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In all of these cases, we would be asking the probability of something happening.
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The study of probability and chance is a major area of mathematics.
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It has applications throughout business, science, politics, medicine, and many other fields.
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Being able to know how probable an event is, how likely something is to happen, is extremely important for a huge number of things.
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In this lesson, we will go over some of the basic concepts of probability.
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Still, even though they are basic concepts, it is going to be a really interesting amount of stuff that is going to let us see some really interesting results.
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There is a lot of stuff here, even at the basic level.
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Furthermore, basic probability questions pop up a lot on standardized tests, like the SAT and other standardized tests.
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So, if you are planning on taking any of those in the near future, this is especially useful,
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because you are almost certainly going to see some questions that are exactly like what we are working on here, but probably even easier.
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And while it is not absolutely necessary to have watched the previous two lessons before watching this one,
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we are going to draw very heavily on the previous two lessons in this lesson.
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So, I would really recommend watching those first, if you haven't watched them.
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It is not absolutely necessary, but it will make things a little clearer.
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All right, let's go: the first thing that we want to define is the idea of a sample space.
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In the lesson on counting, we defined the term event to simply mean something happening.
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For example, if we had a coin, and we flipped it, we might name an event E that is the event of the coin coming up heads.
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Of course, there might be possibilities other than the event occurring.
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It might come up something other than heads.
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We call the set of all possible outcomes, everything that could happen when we do something, the **sample space**.
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If we want, we can denote the sample space with a symbol, such as S.
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In the example above, there are two possible outcomes for the sample space.
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1) It comes up heads, or 2) it comes up tails.
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There are two different possibilities, because there are two different sides to the coin.
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Two sides to the coin means two things in our sample space.
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The event is simply heads coming up; the sample space is everything that could occur, heads and tails.
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Probability of an event: let E be an event, and S be the corresponding sample space.
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Let n(E) denote the number of ways that E can occur, and n(S) denote the total possible number of outcomes.
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Then, if all of the possible outcomes in S are equally likely, the probability of event E occurring,
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denoted p(E), probability of E, is p(E) = n(E)/n(S).
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Equivalently, just using words, that is: the probability of an event is equal to the number of ways the event can occur, divided by the number of possible outcomes.
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For example, in a standard 52-card deck, there are 4 of each card in various suits.
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If we draw a card at random from the deck, the chance of drawing an ace is...
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The probability of the event of an ace coming up is: there are four ways that we can get an ace out of the deck.
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There are four aces in there, so there are four ways that that can happen.
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And there are a total of 52 cards in the deck, so there is a total possibility...the total number of possible outcomes
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is pulling any of those 52; so it is 4/52, which simplifies to 1/13.
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This idea is the basic idea of probability: being able to say the number of ways it could occur, divided by the number of all possible things that could happen.
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That is pretty much the main idea; and if you take one idea away from this lesson, this is the one idea to take away.
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And this is the sort of thing that you will end up seeing on any standardized test.
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This idea right here is enough for any standardized test.
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So, as long as you keep that one in your mind, you will be good for all of that.
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We are going to get into slightly more interesting things as we keep going, but this is the basic, fundamental idea that you want to hold onto.
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Equally likely is important: that is a really important phrase.
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In that previous definition, it was assumed that all possible outcomes were equally likely.
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This is a really important requirement for the way that we are going to look at probability.
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Why? Well, let's consider the following scenario.
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If you dig a hole in the ground, there is a possibility that you will find gold.
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The sample space, then, has two possibilities: you find gold, or you don't find gold.
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We have an event, which is finding gold; and our sample space is finding gold or not finding gold.
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That means one thing for the event is divided by two things for the size of our sample space.
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But clearly, if you dig some random hole, the chance of you finding gold is not 1/2.
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Why? Because each chance is not equally likely.
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Each outcome is not equally as likely as the other one.
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Finding gold is not equal occurrence, equal probability, with not finding gold; they are not equally likely outcomes.
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So, because they are not equally likely outcomes, we can't base it off of this idea of
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number of ways of our event, divided by number of possible ways anything could happen.
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All right, the method that we just talked about for that basic probability thing won't work if we don't have this "equally likely" thing.
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This "equally likely" thing is a really important first requirement.
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Now, happily, we are almost certainly never going to see anything that doesn't involve equal likelihood.
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All of the problems that we are going to end up seeing are going to be equal likelihood problems.
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We are going to know that all of the outcomes are equally likely.
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They might describe it with words like fair, like "a fair die" or "a fair coin" or a "fair" random number--
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something that implies that all of the possibilities are equally likely.
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A fair die is one that is equally likely to come up 1, 2, 3, 4, 5, 6.
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All of its outcomes are equally likely.
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We can also use the word "random"; if something is selected randomly, that is implying that, out of the selection, it was equally likely amongst all of them.
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Or some other way of saying this...but we can almost always assume that all of the possible outcomes are equally likely.
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At the level of the problems that we are going to be working at, this is a pretty reasonable assumption to make:
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that we can assume this "equally likely" thing at the level of the problems we are working at.
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There are lots of really interesting problems you can work on, where this assumption won't end up being true.
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But for the sort of thing you will be required to work on at this point,
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you can almost always be certain that you will be allowed to assume
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that everything is equally likely, unless they very explicitly tell you otherwise.
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And that is not going to happen very often, if at all.
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All right, interpreting probability: notice that the probability of an event, p(E) = n(E)/n(S), is always less than or equal to 1.
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This is because n(E) ≤ n(S); n(E), the number of ways the event can occur,
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is always less than or equal to the total number of things that can happen,
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because all of the ways that the event can occur are all inside of the ways that anything could happen.
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E is always contained within our sample space; the event is always contained within the sample space,
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just like heads was contained inside of heads and tails for the sample space.
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The event is always contained inside of the sample space; so the number of ways the event can happen is always going to be smaller,
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or equal to, the number of things in the sample space, total.
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OK, we have this number, then, that can be somewhere between 0 and 1.
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The smallest E could be is 0 ways total; so we are somewhere between 0 and 1.
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How do we interpret this value?
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We interpret it like this, where we have this value that can range between 0 and 1.
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And at 0, it is absolutely impossible for the thing to occur; and at 1, it is absolutely certain that the thing will occur.
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It will definitely occur at a 1 probability, and it will never occur at a 0 probability.
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So, as we end up going up the scale, as we go up from 0 to 1, it becomes more and more likely.
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The closer you get to 1, the more likely it is.
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And here in the middle, at 0.5, it is equally likely as unlikely.
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On average, 1 out of 2 times it will end up happening.
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We can represent probability as a fraction, as a decimal, and as a percentage.
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Any of these are fine things to do; the important part is that probability is always between 0 and 1, inclusive,
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because you can be 0, and you can be 1, as a probability,
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although almost all of the ones we are going to deal with will be somewhere in between.
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We can also interpret probability as the ratio of the event happening over a large number of attempts.
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For example, if we flip a coin a million times, we can expect about half of the flips to come out heads,
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because the probability of flipping a fair coin and having it come out heads is 1 out of 2, 1/2.
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On the large scale, we know that about half of any large thing will end up coming out to be that.
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Now, on the small scale, if I flip a coin twice, it wouldn't be that surprising for two of them to come out as tails,
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even though it is a 1-out-of-2 chance for heads.
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We could flip the coin three times, and it wouldn't be that surprising for it to come out as tails, tails, tails.
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It is not super likely, but it is not that unreasonable.
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Just because the probability is 1/2 for heads doesn't guarantee us that it is going to occur any time.
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With probability, we don't have a guarantee of occurrence; we just have "likely" that it will occur at certain levels of likelihood.
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We can only have certainty at a 1.
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So, since it is a question of how likely it is, we won't have it be as likely to show up, unless we look at a larger sample space.
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We look at a larger number of things that could happen.
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If we look at it happening a million times, we can be almost sure to have half of them be heads,
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because we have done it so many times that we start to see this happen more and more.
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On a very small scale, though, we can't be certain that it will end up showing up.
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We flip a coin three times; it might come up tails all three times, but that doesn't imply that the coin isn't fair.
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It is just how random chance works.
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All right, one of multiple events occurring; consider if we wanted to find the probability of rolling a fair 6-sided die and having it come up either 1 or 6.
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Now, we could consider them as separate events;
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so we would call them E₁ and E₆, the event of a 1 and the event of a 6.
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We can talk about either E₁ or E₆ occurring with the notation E₁ union E₆.
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That is a way of saying E₁ or E₆ or both.
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What we are looking for is something that happens, which is inside of E₁ or inside of E₆.
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We can be inside of either of them, so it is a union.
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What we are curious to know here: we are looking for the probability of rolling a 1 or a 6.
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We are looking for the probability of E₁ union E₆, the probability of E₁ or E₆ or both of them.
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Notice that E₁ and E₆ are mutually exclusive events.
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If one of them occurs, the other one cannot occur.
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What this means is that, if we roll a 1, it is impossible to have rolled a 6 just then.
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If we rolled a 6, it is impossible to have rolled a 1 just then.
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There is no overlap between them; we can't be a 1 and a 6 simultaneously.
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So, they are mutually exclusive events.
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With this in mind, we see that the probability of E₁ or E₆ occurring, E₁ union E₆, just combines their probabilities.
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The probability of E₁ union E₆ is equal to the probability of E₁, plus the probability of E₆.
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Because we don't have to worry about them overlapping, it is just "did E₁ happen?" and then we also could look at "did E₆ happen?"
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The probability of E₁ is 1 out of 6; the probability of E₆, rolling a 6, is 1 out of 6.
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We add those two together, and we get 1/3 as the total probability for what we would have of rolling a 1 or a 6.
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This idea works in general: given two mutually exclusive events, A and B,
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the probability of either one, or both, occurring, A union B, is given by:
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the probability of A union B is equal to the probability of A, plus the probability of B.
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Given that two events can't both happen simultaneously, if we know that, if you are A, then you can't be B,
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and if you are B, then you can't be A, then if we are looking for either one of them happening,
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it is just going to be adding the two probabilities together.
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This is another one of those basic ones that you end up seeing on tests and homework.
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This is a good one to remember.
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What if the events are not mutually exclusive--there is some overlap in the events, so that you could be in A and B at the same time?
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For example, let's consider the probability of a fair die coming up strictly below 4 and/or (so it can also be) coming up even.
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So, the die comes up below 4 (that is 1, 2, 3), or the die comes up even.
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Notice: there is some overlap in these two events.
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The die could come up as 2, which is below 4, and also which is even.
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It is both of these at the same time; it is both of the things.
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Since it is both of the things, there is overlap between being below 4 and being even.
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To find the probability, we have to take this overlap into account.
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In two events, A and B, we denote the overlap of both occurring at the same time as
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A intersect B, that is to say, where A and B are happening at the same time.
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So, where A intersects with B is their area of overlap, where both things are happening at the same time.
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We take the possibility of overlap into account as follows: let A and B be two events.
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The probability of A or B or both is given by probability of A union B, that is to say, A or B, or both occurring,
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is equal to the probability of A, plus the probability of B, minus the probability of A intersect B.
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That is, minus the probability of A and B occurring at the same time.
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So, this is an interesting formula; but that one that we just talked about, where we assumed that they are mutually exclusive--
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that one is much more likely to come up in homework and tests.
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You might end up seeing the formula that we are working on right now.
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But it is less likely; so if this one doesn't make quite as much sense to you, don't worry about it too much.
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We are about to see a quick example, though, that will help explain it.
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So, that might help cement it; but don't worry about it too much if this one doesn't make a lot of sense.
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You are much less likely to see it than the previous one.
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Our previous example, this example that we are looking for, 1, 2, or 3, or an even number, or both of them:
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we can let E<font size="-6">123</font>, which is the event of rolling a number strictly below 4,
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which is to say a 1, a 2, or a 3; and then E<font size="-6">even</font>, which is rolling an even number (you roll a 2, a 4, or a 6)...
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now notice where E<font size="-6">123</font> and E<font size="-6">even</font> overlap.
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These two things overlap at E₂, when we roll a 2.
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So, if a 2 comes up on the die, it is below 4, and it is even.
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So, E₂ is where they overlap each other.
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By the above formula, the probability of A union B, the probability of E<font size="-6">123</font> union E<font size="-6">even</font>,
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of having it be either below 4 or even or both, is equal to the probability of the die coming up as 123,
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plus the probability of the die coming up even, minus the probability of the die coming up both of these
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(the probability of it being 123 and even).
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So, what is the probability of it coming up as 123?
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Well, 1, 2, 3...that is 3 possibilities, divided by 6 total on the thing, so it is 3/6.
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Plus coming up even: 2, 4, 6; that is 3 possibilities, divided by 6 total, so 3/6 as well.
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Minus...and now we can swap, since we know that E₂ is the same thing as E<font size="-6">123</font> intersect E<font size="-6">even</font>.
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So, it is the same thing as just asking what the probability is of rolling a 2.
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We combine red and blue together, and they come to be a 1; 3/6 + 3/6 is 1.
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And then minus...what is the probability of rolling a 2?
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Well, that is 1 (rolling a 2), divided by 6; so we have 1/6; 1 - 1/6 comes out to be 5/6.
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So, we have 5/6 as the chance of rolling a number that is below 4, or even, or both.
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Now, 5/6 makes sense, because we can also just go through this and do this by hand.
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1, 2, 3, 4, 5, 6: notice: if you are 1, 2, or 3, then these ones are good; if you are even: 2, 4, and 6--then those ones are good, as well.
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So, the only one that fails to be 1, 2, 3, or even is the number 5; that means we have
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1, 2, 3, 4...1 here; 2 here; 3 here; 4 here; 5 possibilities for this event to occur,
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divided by a total of 6 possible outcomes; so 5 divided by 6 is the exact same thing, so this checks out and makes sense.
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So, this formula here makes sense on the small scale, and we can also bring it to a much larger scale,
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if we are working with a much more complicated problem, where we can't just do this by hand,
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and we have to be able to understand the theory to be able to get an answer.
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All right, what if an event doesn't occur?
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If we have some event E, we can also talk about the event of E not occurring.
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We have E that occurs when the event E occurs; but we can also talk about if E does not occur; let's make that an event.
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E not occurring is now an event, as well; we call this the complement of E.
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The complement of an event is that event not occurring, and we denote it with E with a little c in the top corner,
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so E to the c, but not actually raising it like an exponent; it is not the same thing...E complement.
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Other textbooks or teachers might denote this as E with a bar on top or E with a little tick mark.
00:17:46.000 --> 00:17:50.100
It doesn't matter; any of them is fine; but I am going to use E^c.
00:17:50.100 --> 00:17:57.100
For example, if E is the event of rolling an even number on a die, then E^c is the opposite event.
00:17:57.100 --> 00:18:02.600
In this case, we have that E is an even number; so E^c would be the complement,
00:18:02.600 --> 00:18:09.100
when E does not occur; so if E does not occur, what is the opposite to rolling an even number?
00:18:09.100 --> 00:18:15.800
That is rolling an odd number; so if you roll an odd number on the die, you are in the complement of the event here.
00:18:15.800 --> 00:18:22.500
The probability of an event's complement occurring is 1 - the probability of the original event.
00:18:22.500 --> 00:18:29.200
So, the probability of an event's complement is equal to 1 minus the probability of the original event.
00:18:29.200 --> 00:18:37.700
Why? Well, either E occurs, or it does not; and if it does not, then we know that E^c must occur.
00:18:37.700 --> 00:18:44.500
So, E^c must occur if we end up having that E does not occur.
00:18:44.500 --> 00:18:49.900
If E occurs, then E has occurred; if E does not occur, then E^c has occurred.
00:18:49.900 --> 00:18:54.400
So, no matter what, we can be certain that one of them must occur.
00:18:54.400 --> 00:18:57.100
One of these two things always has to happen.
00:18:57.100 --> 00:19:01.300
We can't have something in the middle: either the event happens, or the event doesn't happen.
00:19:01.300 --> 00:19:06.300
So, if the event happens, or it doesn't happen, well, either way, one of those two things happened.
00:19:06.300 --> 00:19:09.800
So, we are certain that one of them will occur.
00:19:09.800 --> 00:19:16.000
Since one of them always has to occur, that means that the total of their probabilities must be a 1, certainty.
00:19:16.000 --> 00:19:20.500
The probability of an event's complement, plus the probability of the event, is equal to 1,
00:19:20.500 --> 00:19:24.200
because one of those two things always has to happen.
00:19:24.200 --> 00:19:29.800
So, that is why we end up having 1 minus the probability of the event give us the probability of the event not happening.
00:19:29.800 --> 00:19:31.900
This is a useful idea in a lot of situations.
00:19:31.900 --> 00:19:35.200
So, this is another useful one to remember.
00:19:35.200 --> 00:19:39.300
Independent events: consider rolling a die and flipping a coin.
00:19:39.300 --> 00:19:45.800
How can we find the probability of the die coming up as a 5 or a 6, and the coin coming up heads?
00:19:45.800 --> 00:19:48.800
To do this, we must consider the probabilities of both events.
00:19:48.800 --> 00:19:53.100
In the example of the above events, we say that they are independent events,
00:19:53.100 --> 00:19:58.500
because rolling the die has no effect on flipping the coin, and flipping the coin has no effect on rolling the die.
00:19:58.500 --> 00:20:02.700
They are separate events, and the outcome of one event does not affect the other.
00:20:02.700 --> 00:20:06.000
We know that they are independent events in this case.
00:20:06.000 --> 00:20:13.200
If they are independent events, given two independent events, A and B, the probability of both events occurring,
00:20:13.200 --> 00:20:22.600
that is to say, the probability of A and B occurring, is equal to the probability of A, times the probability of B.
00:20:22.600 --> 00:20:30.100
We multiply the probabilities of each of them on their own, and that gives us the probability of both of them occurring, if they are independent events.
00:20:30.100 --> 00:20:38.700
In the above example, we would have: the probability of an event of a 5 or 6, and the event of a heads,
00:20:38.700 --> 00:20:44.200
is equal to the probability of the event of a 5 or a 6, times the probability of the event of a heads.
00:20:44.200 --> 00:20:45.900
Well, what is the chance of a 5 or a 6?
00:20:45.900 --> 00:20:53.500
A 5 or a 6 is going to be 2 out of 6; and the probability of a heads is going to be 1 out of 2.
00:20:53.500 --> 00:20:58.700
We multiply these together, and we get 1/6.
00:20:58.700 --> 00:21:05.800
This right here, this idea of multiplying probabilities together, is the other really important idea for this lesson.
00:21:05.800 --> 00:21:12.600
That very, very basic one of the number of ways that the event can happen, divided by total number of things that can occur--
00:21:12.600 --> 00:21:15.600
that is the first really important idea in this lesson, and the second one is:
00:21:15.600 --> 00:21:18.900
if you get independent events, and you want to figure out the chance of both of them occurring,
00:21:18.900 --> 00:21:21.800
you just multiply the probabilities of each of them on their own.
00:21:21.800 --> 00:21:27.200
That one also comes up on tests a lot; that is another very important idea to take away from this.
00:21:27.200 --> 00:21:30.600
What if the events are not independent, though--what if we are looking at a situation
00:21:30.600 --> 00:21:36.100
where they aren't actually going to not affect each other, where they can have an effect on each other?
00:21:36.100 --> 00:21:41.200
The outcome of one will do something to the outcome of the other, or at least the chances of the outcome of the other.
00:21:41.200 --> 00:21:45.900
For example, consider of we draw two cards at random from a deck of 52 cards.
00:21:45.900 --> 00:21:50.200
What is the probability of a pair of aces--that is, both cards coming out as aces?
00:21:50.200 --> 00:21:56.800
In this case, the probability of the second card being an ace is affected by what the first card was,
00:21:56.800 --> 00:21:59.400
because it can change the number of aces in the deck.
00:21:59.400 --> 00:22:05.900
If we pull out a card that is an ace on the first one, then there is now one fewer ace for our second pull.
00:22:05.900 --> 00:22:11.000
If we don't pull out an ace on the first one, then there is the same number of aces in our second pull.
00:22:11.000 --> 00:22:16.300
So, what we do on that first pull affects what will happen in the second pull.
00:22:16.300 --> 00:22:23.500
We denote conditional probability with this sort of symbols, this notation.
00:22:23.500 --> 00:22:31.800
The conditional probability of B occurring if A does occur is p of B bar A n things.
00:22:31.800 --> 00:22:34.200
So, that is how we would say it--just the symbols.
00:22:34.200 --> 00:22:38.000
If we want to talk about it, it would be the conditional probability of B occurring if A occurs.
00:22:38.000 --> 00:22:43.600
Or we could also say this as the probability of B occurring, assuming that A occurs.
00:22:43.600 --> 00:22:51.500
So, if we can know for sure that A will occur, then what is B's chance of occurring, with that piece of information already in mind?
00:22:51.500 --> 00:22:54.100
This is the idea of conditional probability.
00:22:54.100 --> 00:23:11.100
We assume the second one; the second one is the assumed one, and the first one is what we are looking at.
00:23:11.100 --> 00:23:17.900
We assume the second one, p of B, A, the conditional probability of B occurring if A does, so B bar A.
00:23:17.900 --> 00:23:21.300
A is going to be assumed; the second thing that shows up is assumed.
00:23:21.300 --> 00:23:25.200
The first thing is now what we are trying to figure out the probability of, if we can have that assumption.
00:23:25.200 --> 00:23:32.500
All right, so in the above example, we could denote the probability of the second ace being pulled, if the first ace has already been pulled.
00:23:32.500 --> 00:23:37.300
What is the chance of that second ace coming out, if the first ace has already come out?
00:23:37.300 --> 00:23:43.000
Then, we have our assumed thing, the first ace; and what we are looking for now is
00:23:43.000 --> 00:23:50.700
the probability of that second ace, with that assumption there: the probability of the second ace, assuming a first ace.
00:23:50.700 --> 00:23:55.600
Given two events A and B, where the outcome of A affects the outcome of B,
00:23:55.600 --> 00:24:03.400
the probability of both events occurring is: probability of A and B is equal to the probability of A occurring on its own,
00:24:03.400 --> 00:24:10.600
multiplied by the probability of B occurring, assuming that A occurs (this conditional probability).
00:24:10.600 --> 00:24:13.500
Let's see this as an example: in the previous example, the result of one card
00:24:13.500 --> 00:24:17.600
affects the result of the second card, because it changes what is in the deck.
00:24:17.600 --> 00:24:27.700
So, the probability of that very first ace is 4 out of 52, because there are 4 aces in the deck when we pull it out, and there are 52 cards total to pull from.
00:24:27.700 --> 00:24:37.600
But for the second ace, there is now one less ace, and there is one less card, if we assume that an ace was pulled on that first draw.
00:24:37.600 --> 00:24:41.100
So, there is one less ace; that means that we now have 3 aces to pull from,
00:24:41.100 --> 00:24:46.600
divided by 51 cards that we are pulling from total, because now there is just one less card in the deck.
00:24:46.600 --> 00:24:51.900
Thus, the probability for drawing a pair of aces--if we were looking for the probability of both of these happening,
00:24:51.900 --> 00:24:57.900
the first ace and a second ace pulled on two cards, then the probability of that first ace,
00:24:57.900 --> 00:25:05.100
times the probability of the second ace, assuming a first ace has already been pulled, is how we get this.
00:25:05.100 --> 00:25:11.400
We multiply those two together: so the probability of the first ace was 4/52, and then we multiply it
00:25:11.400 --> 00:25:15.900
by the probability of the second ace, if we can assume that the first ace has already happened.
00:25:15.900 --> 00:25:24.100
4/52 times 3/51...that comes out to be 1/221.
00:25:24.100 --> 00:25:27.000
All right, that idea of conditional events is really interesting stuff.
00:25:27.000 --> 00:25:31.700
But it is also probably the most extreme stuff that you would end up seeing on a test.
00:25:31.700 --> 00:25:36.300
I would doubt that you would even see that very often, but it is pretty cool; we will see that in the final example.
00:25:36.300 --> 00:25:42.700
All right, let's start with a nice, simple example: Given a bag containing 4 red marbles, 7 blue, 11 green, and 2 purple,
00:25:42.700 --> 00:25:46.900
what is the chance of a blue marble being drawn randomly from the bag?
00:25:46.900 --> 00:25:52.900
We are looking for what the probability is of a blue coming out.
00:25:52.900 --> 00:25:58.200
How do we figure out probability? Well, we know that we are drawing randomly.
00:25:58.200 --> 00:26:02.900
So, if it is randomly, we know that all of the possibilities are equally likely; that is what that "random" word means:
00:26:02.900 --> 00:26:05.900
that it is a tip that says that all of these are equally likely;
00:26:05.900 --> 00:26:08.800
you don't have to worry about it, which means that we can use that nice, simple formula.
00:26:08.800 --> 00:26:15.300
So, it is the number of ways that the event can occur (in this case, the number of blues we could pull out),
00:26:15.300 --> 00:26:27.500
the number of blue marbles, divided by the number of marbles total, all of the ways that something can happen.
00:26:27.500 --> 00:26:31.900
In this case, we know...how many marbles are there that are blue? 7 are blue.
00:26:31.900 --> 00:26:35.400
So, it is 7 divided by...how many marbles do we have total?
00:26:35.400 --> 00:26:45.600
We have 4 red; we have to include the 7 (they are still 7, so they count as some of them), plus 7, plus 11, plus 2.
00:26:45.600 --> 00:26:55.400
That is all of the marbles totaled together: we work that out: we have 7/(4 + 7 + 11 + 22), or 24; so 7/24.
00:26:55.400 --> 00:26:58.700
We can't simplify it any more; and there is our probability.
00:26:58.700 --> 00:27:04.400
A 7 out of 24 chance exists of pulling a blue marble randomly from the bag.
00:27:04.400 --> 00:27:08.200
The second example: a class has a breakdown in grades shown by the table below.
00:27:08.200 --> 00:27:11.900
If a student is selected randomly from the class (once again, there is that word "randomly";
00:27:11.900 --> 00:27:18.300
it means that we can just assume that everything is equally likely), what is the chance that they have a B or a C?
00:27:18.300 --> 00:27:30.800
If we are doing this, the probability of a B or a C...well, we can also look at that as the union, B union C.
00:27:30.800 --> 00:27:34.500
It can be B, or it can be C; so we can add these together.
00:27:34.500 --> 00:27:42.300
That would be equal to the probability of a B, plus the probability of a C, because it can be either B or C.
00:27:42.300 --> 00:27:47.200
And we know that they are mutually exclusive; we don't have to worry about them being B and C simultaneously.
00:27:47.200 --> 00:27:49.400
So now, we can work this out: what is the probability of a B?
00:27:49.400 --> 00:27:55.400
Well, there are 12 students that have B's; let's expand a little bit more...
00:27:55.400 --> 00:28:00.300
Here is the probability of B's, and then we will have the probability of C's over here.
00:28:00.300 --> 00:28:05.700
How many students have B's? We have 12 students with B's, divided by...how many students do we have total?
00:28:05.700 --> 00:28:12.300
We have 8 here, 12 here, 5 here, 7 here; so we add them all together to figure out how many students we have in the class.
00:28:12.300 --> 00:28:16.900
8 + 12 + 5 + 7; then we add...what is the probability of a C student?
00:28:16.900 --> 00:28:20.900
Well, a C student has 5...there are 5 C students total in the class.
00:28:20.900 --> 00:28:26.900
And then, once again, we divide it by the number of students total, 8 + 12 + 5 + 7.
00:28:26.900 --> 00:28:41.600
We work this out: 12 over...8 + 12 is 20, plus 5 is 25, plus 7 is 32, plus 5/32, equals 17/32.
00:28:41.600 --> 00:28:44.100
We can't simplify that any more, so there is our probability.
00:28:44.100 --> 00:28:53.200
If we draw a student randomly from the class, we have a 17/32 chance of pulling a B or a C student.
00:28:53.200 --> 00:28:58.200
Next, a bag of marbles contains 12 red marbles, along with various others.
00:28:58.200 --> 00:29:03.700
If you draw a marble from the bag at random, you have a 15% chance of drawing a red one.
00:29:03.700 --> 00:29:05.900
What is the total number of marbles in the bag?
00:29:05.900 --> 00:29:12.600
So, for this, we have 12 red marbles; we know that, if we pull out of the bag at random, we have a 15% chance of drawing a red one.
00:29:12.600 --> 00:29:18.600
So, our very first thing is that we want to turn 15% as a chance...we can't really work with percentages.
00:29:18.600 --> 00:29:23.400
We have seen this before; we have to turn them into decimal numbers before we can work with them in math, usually.
00:29:23.400 --> 00:29:28.700
15% we change into 0.15; so it is a 0.15 probability.
00:29:28.700 --> 00:29:35.100
Now, if we work this our normal way, then we know that the probability of pulling out a red
00:29:35.100 --> 00:29:48.300
is equal to the number of reds in the bag, divided by the number of marbles total.
00:29:48.300 --> 00:29:50.400
The total number is on the bottom.
00:29:50.400 --> 00:29:52.200
What is the probability of pulling out a red?
00:29:52.200 --> 00:29:58.500
Well, that is a 0.15; how many red are there? There are 12 red marbles.
00:29:58.500 --> 00:30:05.200
So, it is 12 divided by the number total; the number total is just some number.
00:30:05.200 --> 00:30:08.500
If we wanted, we could replace it with x, or whatever we felt like.
00:30:08.500 --> 00:30:12.900
I am just going to keep writing "number total," because we know that it is just a number that we are working with.
00:30:12.900 --> 00:30:18.400
We multiply both sides by number total; number total will cancel out on the denominator on the right, and appear on the left.
00:30:18.400 --> 00:30:24.400
And we divide both sides by 0.15; so now, we have 12 over 0.15.
00:30:24.400 --> 00:30:30.700
We use a calculator to figure out what is 12 divided by 0.15; that comes out to be 80.
00:30:30.700 --> 00:30:36.300
So, we know that the total number of marbles in the bag is 80 marbles, because we were guaranteed
00:30:36.300 --> 00:30:42.200
that if we pull out randomly from the bag, there is going to be a 15% chance of drawing a red one.
00:30:42.200 --> 00:30:44.500
And we knew how many red marbles we started with.
00:30:44.500 --> 00:30:48.500
It is a probability, with just a slight spin of algebra on it.
00:30:48.500 --> 00:30:55.100
The fourth example: If you roll three fair six-sided dice ("fair" means that all of the sides are equally likely),
00:30:55.100 --> 00:31:02.400
what is the chance of a 6 coming up on each die? on none of the dice? and finally, on at least one of the dice?
00:31:02.400 --> 00:31:06.200
So first, let's say we are looking at all dice.
00:31:06.200 --> 00:31:14.700
6 on all: if it is going to be a 6 on all of them, then it is a question of what the first one is.
00:31:14.700 --> 00:31:18.500
Well, the second one is an independent event of the first one.
00:31:18.500 --> 00:31:21.400
And the third one is an independent event of the first two.
00:31:21.400 --> 00:31:25.500
One die's result does not affect the result of the next die, which does not affect the result of the next die.
00:31:25.500 --> 00:31:32.700
They are independent events; so what is the chance of one die coming up as a 6? It is 1 out of 6.
00:31:32.700 --> 00:31:35.900
What is the chance of the next die coming up as a 6? It is 1 out of 6.
00:31:35.900 --> 00:31:39.600
The chance of the next die coming up as a 6 is 1 out of 6.
00:31:39.600 --> 00:31:43.500
By our rule about independent events, it is multiplying them together.
00:31:43.500 --> 00:31:49.000
If they are independent events, and you want to know what the probability of them all occurring is, just multiply all of the probabilities together.
00:31:49.000 --> 00:32:01.000
1/6 times 1/6 times 1/6...we work that out; that comes out to 1/216; so it is a 1 out of 216 chance of a getting a 6 on all of the dice.
00:32:01.000 --> 00:32:06.200
Next, what if we want to do none of the dice?
00:32:06.200 --> 00:32:13.600
If it is none of the dice, what is the chance of getting no 6's on any die?
00:32:13.600 --> 00:32:17.200
Well, that would be 1, 2, 3, 4, 5...those are the things that are allowed to come up.
00:32:17.200 --> 00:32:22.500
So, we have 5 possibilities, divided by 6, the total number of things that could happen.
00:32:22.500 --> 00:32:28.800
And that is going to be the same for the second die and the third die: 5/6 times 5/6 times 5/6.
00:32:28.800 --> 00:32:32.800
We could also write this as 5³, divided by 6³.
00:32:32.800 --> 00:32:41.200
We work that out with a calculator, and we get 125...actually, we probably don't need a calculator for that...divided by 216.
00:32:41.200 --> 00:32:48.400
So, the chance of getting no 6's on any dice is 125/216.
00:32:48.400 --> 00:32:54.200
Finally, at least one of the dice: this might be the one that seems hardest at first,
00:32:54.200 --> 00:32:59.500
but it is actually easy, once we know this part right here, if we know none of them.
00:32:59.500 --> 00:33:02.600
So, remember what we talked about before with the complement of an event.
00:33:02.600 --> 00:33:08.200
If an event does not happen, then we are talking about the event complement happening.
00:33:08.200 --> 00:33:12.200
If E does not happen, then E^c does happen.
00:33:12.200 --> 00:33:16.400
If none of the dice happens, then there are no 6's on any of them.
00:33:16.400 --> 00:33:21.900
But if none of the dice does not happen--that is to say, we don't roll a 6 on none of the dice--
00:33:21.900 --> 00:33:25.000
then that means we have rolled a 6 on one of the dice, or more.
00:33:25.000 --> 00:33:30.400
So, at least one is going to be the probability of the complement to the none event.
00:33:30.400 --> 00:33:42.800
So, what we talked about before was the probability of none, complement: it is equal to 1 - the probability of none.
00:33:42.800 --> 00:33:47.000
Another way of looking at it is just that, if you know what the probability of an event is,
00:33:47.000 --> 00:33:53.400
then the probability of the opposite thing happening is 1 minus the probability of that event; that is what this none complement thing is.
00:33:53.400 --> 00:33:55.700
It is the opposite of that event occurring.
00:33:55.700 --> 00:34:03.400
We know 1 minus...we just figured out what is the probability of none occurring; it is 125 over 216.
00:34:03.400 --> 00:34:14.300
So, we get that 91 out of 216 is the chance that at least one of the dice will come up with a 6 on it.
00:34:14.300 --> 00:34:18.800
The final example: A poetry class has 17 boys and 13 girls.
00:34:18.800 --> 00:34:25.500
If the teacher randomly selects 4 students from the class, what is the chance that they will all be boys?
00:34:25.500 --> 00:34:28.600
This is the idea of conditional probability.
00:34:28.600 --> 00:34:36.600
We have some first thing that is going to happen; but then, the second thing is going to also be affected by that first thing.
00:34:36.600 --> 00:34:40.900
And then, the third thing is going to be affected by that second, and the first, thing.
00:34:40.900 --> 00:34:45.700
And then, the fourth thing is going to be affected by that first, second, third thing as well.
00:34:45.700 --> 00:34:51.200
So, what we do is figure out the probability of the first thing; then we multiply it by the probability of the second thing, assuming that first thing.
00:34:51.200 --> 00:34:54.400
And then, we multiply that by the probability of the third thing, assuming those first two things.
00:34:54.400 --> 00:34:57.700
And then, we multiply that by the probability of the fourth thing, assuming those first three things.
00:34:57.700 --> 00:35:01.400
That is how conditional probability worked when we talked about it earlier.
00:35:01.400 --> 00:35:10.300
So, if we start with how many students there are total in the class...if we have 17 boys and 13 girls, we assume we have 30 total.
00:35:10.300 --> 00:35:16.600
So, if we have 30 total, then for the first one, we have 17 out of 30.
00:35:16.600 --> 00:35:20.000
But for the next one, we pull out one of the students.
00:35:20.000 --> 00:35:23.800
We have pulled out one boy; so that means we now only have 16 boys.
00:35:23.800 --> 00:35:30.100
How many students do we have total? Our total of students has also gone down by 1, because we have already used one of the students.
00:35:30.100 --> 00:35:36.100
We pulled out a boy; he is still one of the students in the class, so we now reduce from 30 to 29 students in the class.
00:35:36.100 --> 00:35:42.600
Next, we pull out another boy; we are going to now be at 15 boys, divided by...we pull out another student...28.
00:35:42.600 --> 00:35:50.800
Finally, our fourth boy: we are now at 14 boys left to pull from, and we are now at 27 students total in the classroom, after our three pulls so far.
00:35:50.800 --> 00:36:05.200
If we want to figure this out, conditional probability is that we just multiply them all together: 17/30 times 16/29 times 15/28 times 14/27.
00:36:05.200 --> 00:36:17.800
We work that out, and that ends up simplifying to the not-that-simple-looking 68/783, which comes out approximately to 0.087.
00:36:17.800 --> 00:36:26.000
We have a little bit less than a 10% chance, a .087 chance, of managing to pull all boys if we pull 4 students.
00:36:26.000 --> 00:36:32.200
It drops down pretty quickly; the first boy has a 17/30 chance, but we drop down pretty quickly by the time we are at the fourth boy.
00:36:32.200 --> 00:36:34.600
It is a less-than-one-in-ten chance.
00:36:34.600 --> 00:36:41.400
All right, that finishes all of our stuff about combinatorics in here, our ideas of counting, permutations, combinations, and probability.
00:36:41.400 --> 00:36:44.600
I hope you have a reasonable grounding; this is all of the basics that you need for this level of math.
00:36:44.600 --> 00:36:47.400
But there is a huge, huge amount of stuff to explore out there.
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If you thought that this stuff was interesting, just do a quick search on combinatorics.
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You will find out all sorts of cool things; there are all sorts of really cool things in combinatorics--
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how to count things; there are lots of cool ideas in that.
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All right, we will see you at Educator.com for the next lesson--goodbye!