WEBVTT mathematics/probability/murray
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Hi, welcome to www.eudcator.com, my name is Will Murray and
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we are going to be starting the probability lectures today.
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Thank you very much for joining us, we are going to jump right on in.
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We are going to learn some terminology in this first lesson.
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We are going to learn about experiments and outcomes and sample spaces and events.
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Let us get right started with that.
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The words that we are throwing at you are all used to describe probability experiments.
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Let me first got introduce the word experiment itself.
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An experiment is a process leading to exactly one of various possible outcomes.
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We will see lots of examples of this as we go through the lectures.
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But some example to keep in mind, a very simple example would be flipping a coin, or you roll a dice,
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and you draw a card from a deck.
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Those are the kinds of things that we can run in an experiment in exactly one of the possible outcomes will occur.
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Having said that, I should say what an outcome is one of the things that can happen in an experiment.
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Outcomes are also known as simple events or sample points.
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All of those words are synonymous, we use them interchangeably.
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A simple event, a sample point, or outcome.
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If you put all the possible outcomes together that is called the sample space.
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The sample space is essentially the set of all things that can happen in an experiment.
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If you are going to flip one coin, for example the sample space is head and tail
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because those are the 2 things that can happen.
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You can get a head or a tail.
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The sample space is also known as probability space.
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Those words were used interchangeably, probability space or sample space.
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If you are a studying probability, with your own textbook they might use the word sample space,
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they might use the word probability space, those mean the same thing.
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Finally, an event is a subset of the sample space that is a set of some of the outcomes.
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An event means for example, if we are going to pick a card from a deck of cards,
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the event could be that we get a spade because there are 52 cards in a standard deck and 13 of them are spades.
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We collect all those 30 outcomes together and call them an event.
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We often use the variable S to describe the set of all possible outcomes so the sample space.
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We often use the variable A or sometimes A and B when we have multiple ones to describe different events in that sample space.
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We will talk about the probability of an event A occurring when we run an experiment.
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Let us figure out how we compute the probabilities.
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When we do that, if we have an event is assuming that these are things that we can count and everything is nice and finite,
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first of all we count the number of outcomes in the sample space S.
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We count all the possible things that can happen in an experiment.
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If you are flipping a coin, the only 2 things that can happen are head and tail.
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You just count the head and the tail, you get 2 possible outcomes.
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If you are drawing a card from a deck, there are 52 different things that can happen
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because you can get the ace of spades or the 2 spades, up to the king of spades and so on.
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There are 52 different cards that you might draw.
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Those are all the outcomes.
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To calculate the probability of an event, you want to count the number of outcomes in that event.
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For example, if you are drawing a card from a 52 card deck and
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you want to calculate the probability of getting a spade, P of getting a spade when you draw a single card from a deck.
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There are 52 possible outcomes, there are 52 possible cards that you can draw, and 13 of them qualify as being in that event.
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There are 13 possible outcomes which will give you spades.
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Of course, you can reduce the fraction 13/52 and you find that the probability of getting a spade is ¼.
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That is not very surprising.
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Because of this of nature a probability, you are counting things all over the place.
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There is a lot about counting outcomes, counting events.
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A great deal of early focus in probability course is it is going to have can have a common notarial flavor,
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when we try to count different events and count different sample spaces.
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You will see that some of the problems that we study today and in some of the other early lectures
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are all about counting different things, different complicated counting techniques.
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You get the hang of it as we get into it, as we start practice it a little more.
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Our first example is to roll 2 dice and to see if the sum showing on the 2 dice is 10.
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Remember, these are standard dice so they are labeled 1 through 6 on the 6 sides.
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I want to see if the sum is going to be a 10 between the 2 dice.
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We want to identify the experiment.
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We want to list all the possible outcomes and the event that we are interested in,
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which means the subset of outcomes that would give us 10.
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We want to find the probability of the event.
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The probability that we are going to get a 10 when we roll 2 dice.
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This is a pretty standard probability problem.
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The experiment here is just roll the 2 dice.
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We are rolling the dice, roll the 2 dice.
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The outcomes are all the different possible ways that the 2 dice can call up.
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And it is very helpful here if you have a mental image of the dice being distinct.
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Maybe think of one of them as being red and one of them as being blue.
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Think of you are rolling a red dice and a blue dice.
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Let us think of all the different combinations that can arise there.
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The red dye could be 1, 2, 3, 4, 5, and 6.
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The blue dice could be 1, 2, 3, 4, 5, and 6.
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They are all in the different possible combinations of the 2 dice there.
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The red could be 1, the blue could be 1.
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The red could be 1 and the blue could be 2.
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The red could be 1 and the blue could be 3, and so on.
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There are 6 possible combinations in that first roll there.
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There are going to be 6 possible combinations in the 2nd roll, 3rd roll, 4th roll, 5th roll, all the way up to
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you could get 6 on both sides.
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You can get red dice being 6 and the blue dice being 6.
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There are 36 possible outcomes in the set there.
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Our sample space here is all those possible outcomes.
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We can have a red 1 and a blue 1.
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We can have a red 1 and a blue 2.
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Let me see if I can make it look more blue there.
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All the way up to a red 6 and a blue 6.
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We have 36 possible outcomes in this experiment, 36 possible combinations of numbers on the 2 dice.
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That is our sample space, all the possible outcomes.
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We want to identify the event we are interested in.
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The event here is that the 2 dice sum up to 10.
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Let us think about which of those possible outcomes will give us a sum of 10.
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You can get a red 4 and a blue 6.
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You can get a red 5 and a blue 5, because those that up to 10.
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Or you can get a red 6 and a blue 4, because those also add up to 10.
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3 of those combinations between red and blue add up to 10.
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That is the event we are interested in.
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We now know what our event is and we now know what our sample space is.
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Let us figure out what the probability is.
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The probability of event A is the number of outcomes in A ÷ the total number of outcomes in our sample space.
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The total number of things that can happen, outcomes in S.
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And we calculated both of those, we looked at event A and we saw that there were 3 outcomes that will give you a sum of 10.
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In S, there are 36 possible outcomes and that simplifies down into 1/12.
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That is the probability of rolling a 10, when you roll a 2 dice together.
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That is our final probability there.
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Just to recap here, what we did was we identified our experiment which was rolling the 2 dice.
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We listed all the possible outcomes, all the possible outcomes was this chart right here.
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There are 36 possible outcomes, possible combinations of what might be showing on the red dice,
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what might be showing on the blue dice.
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I did not actually list all 36 because that will be too cumbersome but we know there will be 36 of them.
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And then we have to identify the event which is that 2 dice add up to 10.
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There are 3 possible outcomes, 3 of those combinations add up to 10.
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We mention right here a place that often confuses students when I’m first starting to study probability,
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people wonder that it looks like 4 or 6 got listed twice and 5 -5 only got listed once.
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That is true, that is because there are 2 different ways you can get a 4 – 6.
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You get a red 4 and blue 6 or you can get a red 6 and blue 4.
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That will get listed twice.
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5 and 5, there is only one way you can get it which is to get 5 on both of the dice.
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There really are 2 ways to get a 4 and 6.
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There is only one way to get a 5 and 5.
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In fact, this is even reflected when people roll dice in casinos in the game craps,
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people talk about getting 10 the hard way which means getting a 5 and 5
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because that is harder than getting a 4 and 6 because there are 2 ways to get a 4 and 6.
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We listed all of them, we found 3 outcomes there.
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The probability of our event is just the number of outcomes in the event and the number of outcomes ÷
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the number of outcomes in our sample space.
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There are 3 outcomes in our events, we counted those.
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There are 36 in our sample space, 3/36 reduces down to 1/12.
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Let us keep going.
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In our second example, we are going to flip a coin 3 times and see if we get exactly 2 heads.
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And again, we want to identify the experiment, all the possible outcomes in the event we are interested in.
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We want to find the probability of the event.
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The experiment here is flipping the coin.
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The experiment is to flip the coin 3 times, all the possible outcomes
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or all the possible combinations of heads and tails that we can get.
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The possible outcomes here, I’m just going to try to list them systematically.
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Let me call this S because this is the sample space that we are listing right now.
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S is we can get head, head, head.
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We can get a head, head, and a tail.
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We can get a head, a tail, and then a head.
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A head, then a tail, then another tail.
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We can get a tail-head-head.
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We can get tail-head-tail.
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Tail-tail-head or a tail-tail-tail.
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There are 8 possible outcomes here and that is really coming from the fact that 2 × 2 × 2 is 8.
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Because each time you flip the coin, there are 2 different things that can happen.
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Then you multiply those possibilities as you go along, so you get 8 possible combinations of flips there.
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Those are the outcomes that we are interested in, that is the entire sample space.
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The event that we are interested in is that we get exactly 2 heads.
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I want to figure out which of those outcomes satisfies that criterion.
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Exactly 2 heads would mean certainly HHH, it does not but HHT does and HTH does, head- tail- head.
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Down there I see a tail-head-head
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Those are the only ones there are that have exactly 2 heads.
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Our event just has those 3 possible outcomes that satisfy the requirement.
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And then finally, we want to find the probability of the event.
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The probability of A is the number of outcomes in A ÷ the total number of outcomes
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in our sample space, in our probability space, outcomes in S.
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In A, we had 3 possible outcomes and in S we had 8.
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That does not reduce so our final answer there is 3/8.
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That is the probability of getting exactly 2 heads, when we flip a coin 3 times.
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Let we make sure that all the steps there are clear, we want to identify the experiments.
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Identifying the experiment is just flipping the coin 3 times.
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We wanted to list all the possible outcomes.
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I listed them in order here, I kept track of the first flip, the second flip, and the third flip.
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There are 8 possible outcomes.
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That came from 2³ because there are two things that can happen on the first flip × 2 things
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that can happen on the second flip × 2 things that can happen on the third flip.
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Exactly 2 heads, I just scan through that list of outcomes and I have identified the ones that have exactly 2 heads.
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The first one has 3 heads, that does not count.
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The second one and the third one both have exactly 2 heads.
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The 5th one here has exactly 2 heads.
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And that is why I took those and put them into now the set, describing the event that we are interested in.
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The probability of that event is just the number of outcomes in A which is 3 ÷ the number of outcomes in S which is 8.
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That is where we got the probability being 3/8 there.
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For our next example, we are going to be drawing cards from a deck.
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We are going to take a standard 52 card deck and we are going to keep drawing cards until we find the ace of spades.
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The question is what is the probability that it will take us between 20 and 30 cards,
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including 20 and 30, that is what inclusive means.
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When it says inclusive, it means you include those starting and ending values.
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It says identify the experiment, all the outcomes, and the event that we are interested in.
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The experiment here is drawing the cards.
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We are drawing the cards, do not know how long it is going to last because we might get the ace of spades on the very first pick.
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Or if we get unlucky, we might have to draw 51 cards and then get the ace of spades on the very last pick.
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The set of possible outcomes really depends on where we get the ace of spades.
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We could get the ace of spades on the first pick, if we are very lucky.
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We could get it on the second pick, that would still be pretty lucky if we get the ace of spades on the second pick,
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anywhere up to the 52nd pick.
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It would be our worst case scenario, if we got the ace of spades as the very last card that we draw.
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Remember, we said without replacement which means after we draw a card, we are not putting it back in the deck.
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We know for sure that eventually I’m going to find the ace of spades.
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If we are putting the cards back in the deck, I might potentially never find the ace of spades
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because I could keep drawing forever and just keep drawing new cards.
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The event here is that we are going to draw between 20 and 30 cards.
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Our event would be that we get the ace of spades on the 20th pick or possibly on the 21st pick,
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or all the way up to the 30th pick.
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That is our event there, that is our A.
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Generally, use A and B for events.
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We use the letter S for the sample space.
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The probability of A, that is what we are really trying to calculate here,
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is the number of outcomes in A ÷ the number of outcomes in S.
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We just need to calculate that.
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The number of outcomes in S, we already said there is 1 pick up to 2 picks up to the 52nd pick.
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There are 52 possible outcomes in S.
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If we look at A, 20, 21, up to 30, there are 11 of those outcomes because we are counting the 20 and 30 on both ends.
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It is 20, 21, 22,23, 24, 25, 26, 27, 28, 29, 30.
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It is 11 outcomes in the event A.
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We actually have 11 outcomes not 10.
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It seems like there are 10 but there is extra 1 on the end.
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That is our probability of drawing the ace of spades somewhere between the 20th card and the 30th card.
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Let me go back over that and make sure everything is clear.
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The experiment here is drawing the cards.
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We are drawing the cards one by one.
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We might just draw a single card.
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We might end up drawing all the way to the 52nd card because we stop whenever we get the ace of spades.
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And we are not putting cards back in the deck, that is the very key point to mention here.
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The outcomes are that we could get the ace of spades on the first pick, the second pick, all the way up the 52nd pick.
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That is 52 possible outcomes in our sample space.
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The event that we are interested in, according to the problem is that we might get the ace of spades
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on the 20th pick, the 21st pick, or all the way up through the 30th pick.
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There are 11 of those possibilities that we are interested in to satisfy the requirements of the problem.
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When we are calculating the probability, we divide those numbers together.
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The number of outcomes in the event we are interested in ÷ the total number of outcomes just gives us 11/52.
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In example 4, it is the same experiment as before.
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We are going to roll 2 dice, I think we saw that in example 1.
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This time want to see if the sum is a prime number.
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Again, we are going to identify the experiment, all the outcomes in the event we are interested in and
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we are also going to find the probability of the event.
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The experiment, just like before, is to roll 2 dice.
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I think this was in example 1 before.
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Yes, that was example 1.
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Our experiment is the same thing, we are going to roll 2 dice but now the outcomes, those are also the same as before.
00:22:09.000 --> 00:22:12.800
Those are all the possible combinations you can get when you roll 2 dice.
00:22:12.800 --> 00:22:17.500
And again, it is very helpful to think of the dice as being red and blue.
00:22:17.500 --> 00:22:25.000
You can get a 1, 2, 3, 4, 5, and 6 on the blue dice.
00:22:25.000 --> 00:22:32.100
You can also at the same time get 1, 2, 3, 4, 5, and 6, on the red dice.
00:22:32.100 --> 00:22:36.000
In terms of combinations, you can have red 1 blue 1.
00:22:36.000 --> 00:22:41.200
You can have red 1 blue 2, and so on.
00:22:41.200 --> 00:22:55.500
There are 36 possible combinations all the way down to red 6 blue 6.
00:22:55.500 --> 00:23:01.800
That is my set of outcomes, that is my sample space there, 36 possible combinations there.
00:23:01.800 --> 00:23:06.500
This is S and we want to identify the event here.
00:23:06.500 --> 00:23:13.100
The event here is the sum showing on the 2 dice is a prime number.
00:23:13.100 --> 00:23:17.000
Let us figure out which of those combinations could give us a prime number.
00:23:17.000 --> 00:23:24.100
I’m going to use a new screen for this because it takes a little bit of calculation.
00:23:24.100 --> 00:23:29.900
The event here is, I’m going to call it A.
00:23:29.900 --> 00:23:40.900
The set of combinations for which the sum is a prime number.
00:23:40.900 --> 00:23:44.000
Let us remember our prime numbers.
00:23:44.000 --> 00:23:51.400
The prime numbers between 2 and 12, because those are the only possible combinations you can get by rolling 2 dice.
00:23:51.400 --> 00:24:00.500
You can get a 2, 2 is prime, 3 is prime, 5 is prime, 7 is prime, and 11 is prime.
00:24:00.500 --> 00:24:04.000
Those are all the possible totals we are looking for.
00:24:04.000 --> 00:24:10.100
I want to figure out which combinations will give me those possible totals.
00:24:10.100 --> 00:24:15.900
To get a 2, I could roll a 1 on the red dice and 1 on the blue dice.
00:24:15.900 --> 00:24:17.300
That will give me a total of 2.
00:24:17.300 --> 00:24:21.100
That is the only possible combination that does give me a total of 2.
00:24:21.100 --> 00:24:28.300
To get a 3, I can get 1 on the red dice, 2 on the blue dice.
00:24:28.300 --> 00:24:34.300
Or I can get a 2 on the red dice and a 1 on the blue dice.
00:24:34.300 --> 00:24:41.800
To get a 5, I could roll a red 1 blue 4.
00:24:41.800 --> 00:24:46.700
I could roll a red 2 blue 3.
00:24:46.700 --> 00:24:52.000
That was supposed to be blue.
00:24:52.000 --> 00:24:58.300
I could roll a red 3 blue 2.
00:24:58.300 --> 00:25:05.600
Or I could roll a red 4 blue 1.
00:25:05.600 --> 00:25:08.800
There are 4 possible combinations that would add up to 5.
00:25:08.800 --> 00:25:16.500
For 7, I could roll a red 1 blue 6.
00:25:16.500 --> 00:25:21.700
I think I’m going to stop switching colors back and forth because it is slowing me down here.
00:25:21.700 --> 00:25:34.900
I can roll a red 1 blue 6, red 2 blue 5, red 3 blue 4, red 4 blue 3, red 5 blue 2, or red 6 blue 1.
00:25:34.900 --> 00:25:39.600
There are 1, 2, 3, 4, 5, 6 combinations that can give me a total of 7.
00:25:39.600 --> 00:25:52.000
To get 11, I could roll a red 5 and a blue 6, or a red 6 and a blue 5.
00:25:52.000 --> 00:25:53.600
Those are the ways that I can get 11.
00:25:53.600 --> 00:25:56.100
There is a very nice way to keep track of these.
00:25:56.100 --> 00:26:00.000
Instead of trying to list all of these and not being really sure if you got them all,
00:26:00.000 --> 00:26:04.100
let me remind you of what the sample space look quite.
00:26:04.100 --> 00:26:09.900
For blue, we can get 1, 2, 3, 4, 5, 6.
00:26:09.900 --> 00:26:18.400
For red, we can get 1, 2, 3, 4, 5, and 6.
00:26:18.400 --> 00:26:22.300
For red, we can get a red 1 blue 1.
00:26:22.300 --> 00:26:27.500
We can get a red 2, blue 1.
00:26:27.500 --> 00:26:32.500
We can get a red 1 blue 2, and so on.
00:26:32.500 --> 00:26:37.700
We can fill in the rest of this chart but when we are trying to list totals,
00:26:37.700 --> 00:26:44.000
if we look at the total of 2, let me circle that in green here.
00:26:44.000 --> 00:26:48.400
The only way to do it is by getting a 1-1, that is a total of 2.
00:26:48.400 --> 00:26:53.400
If we want a total of 3, we can get 2-1 or a 1-2.
00:26:53.400 --> 00:26:57.300
Those are the ways that you get a total of 3.
00:26:57.300 --> 00:27:08.300
A total of 5 would be, you ca get a red 1 blue 4.
00:27:08.300 --> 00:27:11.600
I’m sorry, I wrote that in the wrong place there.
00:27:11.600 --> 00:27:16.300
That would be a red 4 blue 1.
00:27:16.300 --> 00:27:20.900
You can get a red 3 blue 2.
00:27:20.900 --> 00:27:30.000
You can get red 2 blue 3, or a red 1 blue 4.
00:27:30.000 --> 00:27:40.200
What you get here to get a total of 5 is the diagonal stripe on the table.
00:27:40.200 --> 00:27:46.500
It is 2 stripes below the stripe for 3 because 5 is 2 bigger than 3.
00:27:46.500 --> 00:27:50.100
To get a 7, you will get a diagonal stripe on the table.
00:27:50.100 --> 00:28:00.900
2 places below, you are going to get 6 outcomes across this diagonal stripe.
00:28:00.900 --> 00:28:07.200
There will be 6 outcomes in there, dividing the blue can add up to 6.
00:28:07.200 --> 00:28:19.600
If I need to get 11, you can get a red 6 blue 5 or a red 5 blue 6.
00:28:19.600 --> 00:28:28.100
There is a shorter diagonal stripe there, this is to get 11 and this one is to get 7.
00:28:28.100 --> 00:28:35.300
We can look at these diagonal stripes and the lengths of the stripes to calculate the total probability of this event.
00:28:35.300 --> 00:28:38.000
We just add up all the lengths of those stripes.
00:28:38.000 --> 00:28:44.400
That is kind of a good way to check that we have not overlooked any possibilities here.
00:28:44.400 --> 00:28:47.200
That is our event, let us calculate the probability now.
00:28:47.200 --> 00:29:07.800
The probability of A is equal to the number of outcomes in A ÷ the number of outcomes in S.
00:29:07.800 --> 00:29:11.100
I need to count out the total number of outcomes in A.
00:29:11.100 --> 00:29:14.200
If I look at 2, I see 1 possible outcome.
00:29:14.200 --> 00:29:19.100
If I look at 3, I see 2 possible outcomes.
00:29:19.100 --> 00:29:22.100
5, I see 4 possible outcomes.
00:29:22.100 --> 00:29:24.600
7, I see 6 possible outcomes.
00:29:24.600 --> 00:29:27.100
11, I see 2 possible outcomes.
00:29:27.100 --> 00:29:32.900
Again, I can check these numbers by looking at the lengths of the stripes over here.
00:29:32.900 --> 00:29:35.600
7 has a stripe of 6.
00:29:35.600 --> 00:29:37.600
5 has a stripe of 4.
00:29:37.600 --> 00:29:39.200
3 has a stripe of 2.
00:29:39.200 --> 00:29:40.100
2 has a stripe of 1.
00:29:40.100 --> 00:29:43.400
11 has a stripe of 2.
00:29:43.400 --> 00:29:47.200
Those correspond to these numbers here.
00:29:47.200 --> 00:30:01.000
And if I just add all of those numbers up, I see 2 + 6 is 8 + 4 is 12 + 2 is 14 + 1 is 15.
00:30:01.000 --> 00:30:02.900
I get 15.
00:30:02.900 --> 00:30:09.200
And then remember, the total number of outcomes in S, that is the total size of this chart which was 36.
00:30:09.200 --> 00:30:12.800
It looks like 15 and 36, I can reduce that if I take a 3 out of each one.
00:30:12.800 --> 00:30:18.300
It reduces down to 5/12, that is my probability.
00:30:18.300 --> 00:30:21.900
If I roll 2 dice, what is the probability of getting a prime number?
00:30:21.900 --> 00:30:25.800
The answer is 5 out of 12.
00:30:25.800 --> 00:30:29.900
Let me remind you of everything we did there.
00:30:29.900 --> 00:30:32.900
First, I made a chart of all the possible outcomes.
00:30:32.900 --> 00:30:40.900
I did not fill them all in because it would have taken too long and I got the idea of what the shape would be.
00:30:40.900 --> 00:30:50.000
There are 36 possible outcomes depending on the 6 possibilities for the red dice and the 6 possibilities for the blue dice.
00:30:50.000 --> 00:30:57.400
I identified my event which is that the sum was prime, which means the sum has to be 2, 3, 5, 7.
00:30:57.400 --> 00:31:03.800
I have looked at my chart and I try to list all the outcomes that would lead to the sum being 2, 3, 5, or 7.
00:31:03.800 --> 00:31:06.800
That is where I got 1 -1 for 2.
00:31:06.800 --> 00:31:14.000
2 possibilities adding up to 3, 4 possibilities adding up to 5, 6 possibilities adding up to 7
00:31:14.000 --> 00:31:17.300
and then just 2 possibilities adding up to 11.
00:31:17.300 --> 00:31:20.900
To find the probability, I need to count up all those possibilities.
00:31:20.900 --> 00:31:24.500
I have added up all those numbers, that is what I was doing over here.
00:31:24.500 --> 00:31:28.900
Add them all up to get to 15, that is where this 15 comes from,
00:31:28.900 --> 00:31:33.800
and then divide by the total number of outcomes in the sample space which is 36.
00:31:33.800 --> 00:31:38.500
15/36 reduces down to 5/12.
00:31:38.500 --> 00:31:42.900
That is how we got that probability.
00:31:42.900 --> 00:31:46.200
For example 5, it is a little bit different from some of the others.
00:31:46.200 --> 00:31:50.300
We are going to flip a coin repeatedly until we get a head and
00:31:50.300 --> 00:31:55.000
I want to calculate the probability that we are going to flip at least 3 times.
00:31:55.000 --> 00:31:58.300
The idea is we keep flipping this coin over and over again.
00:31:58.300 --> 00:32:03.600
As soon as we see a head, we stop, and then that is the number of flips that we done in total.
00:32:03.600 --> 00:32:10.200
What is the probability that that number will be 3 or greater?
00:32:10.200 --> 00:32:17.100
Let us identify the experiment and all the outcomes and then find the probability of that event.
00:32:17.100 --> 00:32:31.600
The experiment here is just flipping a coin over and over again.
00:32:31.600 --> 00:32:38.500
And then as soon as we see the first head, we stop.
00:32:38.500 --> 00:32:45.100
The outcomes, this is a little different from all of our example so far
00:32:45.100 --> 00:32:52.800
because what can happen here is we could get a head right away on the first flip.
00:32:52.800 --> 00:32:59.700
Our first possible outcome is we get a head right away or we might get a tail on our first flip
00:32:59.700 --> 00:33:02.500
and then get a head and then would stop.
00:33:02.500 --> 00:33:08.500
Or we might get a tail and then another tail and then get a head and then we would stop.
00:33:08.500 --> 00:33:13.800
We might get 3 tails and then get a head and then we would stop.
00:33:13.800 --> 00:33:17.400
And we do not really know how long this is going to go.
00:33:17.400 --> 00:33:19.800
There is no assuming on how long this is going to go.
00:33:19.800 --> 00:33:22.600
It could go on forever.
00:33:22.600 --> 00:33:26.400
I cannot just list all these outcomes because there are infinitely many.
00:33:26.400 --> 00:33:28.600
It could go on indefinitely.
00:33:28.600 --> 00:33:38.500
The event that we are looking for is that we have at least 3 flips.
00:33:38.500 --> 00:33:45.700
That is A is the set, a single flip getting us a head will be qualified.
00:33:45.700 --> 00:33:47.300
A tail and a head will not qualify.
00:33:47.300 --> 00:33:51.500
It would be a tail- tail-head, that would be 3 flips.
00:33:51.500 --> 00:33:55.900
Tail-tail-tail-head, that would be more than 3 flips.
00:33:55.900 --> 00:34:00.800
A tail-tail-tail-tail-head, that would be more than 3 flips.
00:34:00.800 --> 00:34:04.100
It is anything that has 3 flips or more.
00:34:04.100 --> 00:34:09.400
We want to identify the probability of that event.
00:34:09.400 --> 00:34:12.700
Unfortunately, we cannot do this one by counting.
00:34:12.700 --> 00:34:16.200
Because we have an infinite number of outcomes in our event,
00:34:16.200 --> 00:34:21.200
we have an infinite number of outcomes that can happen from the experiment.
00:34:21.200 --> 00:34:25.100
We can no longer use our counting formula from before.
00:34:25.100 --> 00:34:32.200
Instead, let me introduce a new way of calculating it.
00:34:32.200 --> 00:34:44.000
Our way of calculating the probability is we can add up the probabilities of each of these individual outcomes in our event.
00:34:44.000 --> 00:34:55.900
It is the probability of tail-tail-head + the probability of tail-tail-tail-head + the probability of 4 tails
00:34:55.900 --> 00:35:00.300
and then a head, and so on.
00:35:00.300 --> 00:35:02.200
This is going to be an infinite sum.
00:35:02.200 --> 00:35:06.100
We are going to have to add all these up forever.
00:35:06.100 --> 00:35:08.500
We have to do something clever at some point.
00:35:08.500 --> 00:35:17.400
Tail-tail-head, to get a tail-tail-head, there is a ½ chance that you get a tail right away and
00:35:17.400 --> 00:35:23.100
then there is ½ chance that you get another tail and then there are ½ chance that you get a head.
00:35:23.100 --> 00:35:26.400
Put those together and you get 1/8.
00:35:26.400 --> 00:35:31.700
A tail-tail-tail-head, there are 50% chance of getting a tail, 50% chance of getting another tail,
00:35:31.700 --> 00:35:36.400
50% chance of getting another tail, 50% chance of getting a head after that.
00:35:36.400 --> 00:35:45.800
So it is ½ × ½ × ½ × ½, I you put all of those together you get 1/16.
00:35:45.800 --> 00:35:58.300
Here you will be multiplying together ½⁵ which is ½⁵ is 32.
00:35:58.300 --> 00:36:07.900
We have to add all those up, 1/8 + 1/16 + 1/32.
00:36:07.900 --> 00:36:10.600
This is an infinite series.
00:36:10.600 --> 00:36:13.900
Now, you have to remember something from your calculus 2.
00:36:13.900 --> 00:36:17.500
If you do not remember how to deal with infinite series,
00:36:17.500 --> 00:36:27.500
we got some great lectures here on www.educator.com on the second semester calculus, calculus level 2, taught by a great instructor.
00:36:27.500 --> 00:36:31.000
And there is a lot of work in there in adding up infinite series.
00:36:31.000 --> 00:36:32.400
Here we are going to apply that.
00:36:32.400 --> 00:36:38.200
This is actually a geometric series.
00:36:38.200 --> 00:36:42.600
And with a geometric series, you need to identify the common ratio.
00:36:42.600 --> 00:36:49.300
Here the common ratio, what we are doing from each term to the next is we are multiplying by ½.
00:36:49.300 --> 00:36:52.600
Each term is half as big as the previous term.
00:36:52.600 --> 00:36:59.800
The important thing here is that ½ is less than 1 because that tells us when a geometric series converges,
00:36:59.800 --> 00:37:03.700
it converges when the common ratio is less than 1.
00:37:03.700 --> 00:37:12.000
It does converge and we learn back in those lectures on calculus 2, the sum of an infinite geometric series,
00:37:12.000 --> 00:37:24.700
I’m going to write it in words, it is the first term or 1 - the common ratio.
00:37:24.700 --> 00:37:29.500
In this case, our first term is that 1/8 right there.
00:37:29.500 --> 00:37:35.500
It is 1/8/ 1 - the common ratio which is ½.
00:37:35.500 --> 00:37:41.000
1/8/ 1 -1/2 which is 1/8/ ¼.
00:37:41.000 --> 00:37:48.000
Do a flip here and we get 4/1 × 1/8.
00:37:48.000 --> 00:37:53.500
I’m sorry, 1 - ½ is not ¼, it is ½.
00:37:53.500 --> 00:38:04.000
When we do the flip there, it is 2/1 × 1/8 which is ¼.
00:38:04.000 --> 00:38:14.200
That is our probability of having to flip this coin at least 3 times in order to see the first head.
00:38:14.200 --> 00:38:16.100
I want to go back and calculate this in different way.
00:38:16.100 --> 00:38:18.300
We will see a different way to solve this problem.
00:38:18.300 --> 00:38:19.400
I want to do that on the next page.
00:38:19.400 --> 00:38:25.200
Let me recap how we calculated it using this method first.
00:38:25.200 --> 00:38:29.600
Our experiment was just flipping a coin here.
00:38:29.600 --> 00:38:35.700
The possible outcomes are all the possible strings of heads and tails that we can see.
00:38:35.700 --> 00:38:38.200
Remember, we stop as soon as we see the first head.
00:38:38.200 --> 00:38:40.600
If we get a head right away, we stop.
00:38:40.600 --> 00:38:44.800
If we get a tail, we keep flipping and then we get a head, we stop.
00:38:44.800 --> 00:38:48.100
If we get 2 tails and then a head, we stop after that head.
00:38:48.100 --> 00:38:51.900
If we get 3 tails and then head, we stop after that head, and so on.
00:38:51.900 --> 00:38:54.400
Those are all the possible outcomes.
00:38:54.400 --> 00:38:59.300
The event we are interested in was that we flip at least 3 times.
00:38:59.300 --> 00:39:05.200
I listed all the outcomes that have 3 or more flips.
00:39:05.200 --> 00:39:10.700
To find the probability of that, we cannot count these things anymore because there is infinitely many of them.
00:39:10.700 --> 00:39:15.100
Instead, we have to find the probability of each one.
00:39:15.100 --> 00:39:21.700
The probability of tail-tail-head is 1/8 because it is ½ × ½ × ½.
00:39:21.700 --> 00:39:29.700
Tail-tail-head is 1/16 because you have to predict the flips 4 times in a row to get that right.
00:39:29.700 --> 00:39:34.600
Here, tail-tail-tail-head, you have to predict 5 flips in a row.
00:39:34.600 --> 00:39:38.500
You have 1/2 chance of each one, 1/32.
00:39:38.500 --> 00:39:43.400
Adding all those up, we notice that we get a geometric series.
00:39:43.400 --> 00:39:48.100
Our common ratio, what we are multiplying by each time is ½.
00:39:48.100 --> 00:39:53.100
You got to remember, the sum of a geometric series which we learn back in calculus 2,
00:39:53.100 --> 00:39:59.700
in level 2 calculus, is the first term in the series ÷ 1 - the common ratio.
00:39:59.700 --> 00:40:04.100
That is where I got 1/8 is that first term, that is where that came from.
00:40:04.100 --> 00:40:07.700
The common ratio is that ½.
00:40:07.700 --> 00:40:14.600
1/8/ 1 - ½ simplifies down to ¼ is my probability.
00:40:14.600 --> 00:40:16.800
I’m going to calculate this in a different way.
00:40:16.800 --> 00:40:23.400
Let me show you how you could have done that, in fact without summing up the geometric series.
00:40:23.400 --> 00:40:27.200
Here is an alternate solution for that one, the same one as before.
00:40:27.200 --> 00:40:51.700
Alternate solution is to find the probability of A is equal to, we can calculate the complement of A which means the opposite of A.
00:40:51.700 --> 00:41:00.800
That is the complement A bar means the complement of A, the opposite of what we are looking for.
00:41:00.800 --> 00:41:06.600
And then do 1 - the probability of the complement of A.
00:41:06.600 --> 00:41:08.800
Let us figure out what the complement of A is.
00:41:08.800 --> 00:41:15.700
Remember A was the set of outcomes that have at least 3 flips.
00:41:15.700 --> 00:41:22.800
That was tail-tail-head, tail-tail-tail-head, and so on.
00:41:22.800 --> 00:41:30.100
The complement of A is just the outcomes that head 2 or fewer flips.
00:41:30.100 --> 00:41:35.300
That can be if you get a head right away or you get 1 tail and then a head.
00:41:35.300 --> 00:41:38.900
That is it, because anything else would have 3 or more flips.
00:41:38.900 --> 00:41:47.400
The probability of a complement of A is the probability of getting a head immediately +
00:41:47.400 --> 00:41:51.300
the probability of getting a tail and then a head.
00:41:51.300 --> 00:41:54.900
The probability of getting a head right away on the first flip is ½.
00:41:54.900 --> 00:42:03.700
The probability of getting a tail and then a head is ½ × ½.
00:42:03.700 --> 00:42:11.500
We get ½ + ¼ which is ¾, that was the probability of the complement of A.
00:42:11.500 --> 00:42:19.700
The probability of A is 1 – that, 1 -3/4 which is ¼.
00:42:19.700 --> 00:42:24.400
Of course, that is the same answer we got before but using quite a different strategy.
00:42:24.400 --> 00:42:26.000
It is kind of reassuring.
00:42:26.000 --> 00:42:33.400
They ought to be the same answer but it is nice to see that they check each other.
00:42:33.400 --> 00:42:37.000
Let me make sure that is still clear.
00:42:37.000 --> 00:42:42.500
What we did here was we calculated the compliment of A, that means everything that is not inside A.
00:42:42.500 --> 00:42:45.000
It is the opposite set from A.
00:42:45.000 --> 00:42:51.400
A was tail-tail-head, tail-tail-tail-head, all the strings with 3 flips or more.
00:42:51.400 --> 00:42:57.700
The compliment is all the shorter strings that is just getting a head or getting a tail and a head.
00:42:57.700 --> 00:43:04.800
The probability of that is, the probably just getting a head on the first flip is ½ because you flip 1 coin.
00:43:04.800 --> 00:43:07.300
What is the chance of getting a head is ½.
00:43:07.300 --> 00:43:13.700
The probability of getting a tail and then a head, it would be 1/2 × ½ because you got to get 2 flips,
00:43:13.700 --> 00:43:15.600
the way you are predicting them.
00:43:15.600 --> 00:43:17.500
Add those up, you get ¾.
00:43:17.500 --> 00:43:22.200
We subtract that from 1 because we are using this formula here.
00:43:22.200 --> 00:43:25.300
1 -3/4 is ¼.
00:43:25.300 --> 00:43:34.900
That is the probability of the experiment going for at least 3 flips.
00:43:34.900 --> 00:43:41.200
In our last example here, we are going to draw a 5 card poker hand from a standard 52 card deck.
00:43:41.200 --> 00:43:45.900
The question is what is the probability that we draw all spades?
00:43:45.900 --> 00:43:50.300
We want to identify the experiment, all the outcomes in the event that we are interested in.
00:43:50.300 --> 00:43:53.600
There is a lot of possible outcomes for this experiment.
00:43:53.600 --> 00:44:00.000
We are going to be learning some new notation in the context of this example
00:44:00.000 --> 00:44:04.100
because it is going to introduce something that we are going to be studying later.
00:44:04.100 --> 00:44:08.000
We are going to learn about combinations a little bit in this example.
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The experiment here is drawing the cards, the set of possible outcomes is all the different ways
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to draw 5 cards from a standard 52 card deck.
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This is a little tricky to count.
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Let me show you how it works out.
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Think about it, we are drawing from a 52 card deck, there are 52 ways we can draw the first card.
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After we drawn that first card, there are 51 cards left in the deck.
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There are 51 choices we can make for the second card.
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There are 50 choices left for the third card.
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And then, 49 choices for the 4th and 48 choices for the 5th.
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That seems like there is this very large number of ways to get a particular a poker hand.
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However, once you got the 5 cards in your hand, we do not care what order there in.
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Which means that, you get say the cards you draw are 1, 5, 3, jack, and 8.
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I’m not even going to worry about the difference suits, the clubs, spades, or hearts, or diamonds.
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But suppose you get a 1, 5, 3, jack, or 8.
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You could also had drawn the jack and then the 3 and then the 1 and then the 5 and 8.
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Once those cards are in your hand, that looks like the same poker hand because you can shuffle them around in your hand.
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What we did when we are counting all the different ways to draw the 5 cards
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is we counted these poker hand separately, even though they are really the same poker hand.
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The problem is they were over counting.
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We are counting the same poker hand multiple times.
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We have to take this answer and we have to divide by something and we have to divide by
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the number of ways that each poker hand got counted.
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Now, let us just think about a particular poker hand.
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Let us think about this one in particular.
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1, 5, 3, jack, 8, let us think about the number of different ways that that poker hand was counted in our original list.
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Just think about how many different ways could you order those 5 cards.
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There are 5 choices for which card could have been first, assuming that we are talking about that particular hand.
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There are 5 ways, the 5 choices for the first card.
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Once you have picked that first card, there are 4 choices left for the second,
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and then 3 choices left for the third, and then 2 choices left for the 4th, and then the 5th card would be fixed.
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Remember, we are talking about a particular poker hand like 1, 5, 3, jack, 8.
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We are talking those 5 cards, how many different ways could those 5 cards be rearranged
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and how many different ways that they get counted in this original list.
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The answer is 5 × 4 × 3 × 2 × 1.
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We have to divide that out of our list because each poker hand, each group of 5 cards got counted.
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1 × 2 × 3 × 4 × 5 is 5! ways in our original list.
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We have to divide out by 5! in order to get an accurate number of the different possible poker hands.
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That is the total number of outcomes.
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There is a clever way to simplify this using factorials.
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Notice that the denominator here is 5!.
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The numerator is 52 × 51 × 50 × 49 × 48.
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That looks like we are starting to build up 52!, but what is missing there is 47 × 46, on that down to 1.
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If I just multiply those on, I have to divide them out again.
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47 × 46 on down to 1.
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Another way to write that is as 52! ÷ 47!.
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That is another way to list the total number of hands.
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This is actually something we are going to learn about in the next lesson.
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The total number of ways to choose 5 cards from 52 is, the notation we are going to use is 52 choose 5.
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There are some other notations along with this notation.
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It is a little confusing because different textbooks use different notations.
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C of 52 5, what it means is this factorial formula.
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This is read as 52 choose 5.
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What it means is this factorial formula, you expanded it out by doing 52! ÷ 5!, and this 47!
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really came from the fact that 47 is 52 -5.
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That is where that came from because that 47 came from this 47 here, which came from this 47,
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which came from the fact that we had already used up 5 of the factors of 52! there.
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The number of outcomes here is 52 choose 5.
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This is our new notation, 52 choose 5.
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It is the way you calculate it is as 52! ÷ 5! ÷ 47!.
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Let me recap this because we have to keep going on the next side here.
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I have not even talked about the event that we are interested in yet.
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Our experiment is drawing the cards.
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The number of ways to draw the cards, there are 52 ways to get your first card.
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Once you draw on that card, there is only 51 card left in the deck, so there are 51 ways to get the second card.
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There are 50 cards left in the deck, 50 ways, 49 and 48.
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But the problem is that that over counts because you can get the same poker hand in different possible orders.
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In order to fix that, we looked in a single poker hand, a typical poker hand,
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and we figured out that the number of ways that that could have been ordered was 5 × 4 × 3 × 2 × 1.
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That is why we had to divide out by 5 × 4 × 3 × 2 × 1, it is because each poker hand
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that counted that many ways in our initial list up here.
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Each poker hand, we over counted by a factor of 5!.
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That is our number of outcomes but then as a clever way to write the numerator, we wrote that as 52! ÷ 47!.
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That is where this part of our answer comes from.
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The 47 comes from 52 -5 and then this 5! Is where we got that 5! in the denominator there.
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This is our new notation, 52 choose 5.
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It is sometimes written with these elongated parentheses.
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It is sometimes written with this C.
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C stands for combinations there.
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By the way, this 52 choose 5, it looks like a fraction but it is not a fraction.
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It is not the same as 52/5.
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Do not put the bar there, do not turn it into a fraction because those are totally different things.
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That is just the number of outcomes, we still need to count all the outcomes in the event that we are interested in.
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The event was that we draw all spades.
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Let us try and figure that out.
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I will try and figure that out in the next slide.
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But in fact, it would d be fairly easy if you remember the strategy that we used here.
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Remember that 52 choose 5 is all the ways you can draw 5 cards from 52 cards total.
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Keeping going with this example, our event is A is ways to get a poker hand, a 5 card poker hand of all spades.
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Here is the clever way to think about it.
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Remember how we drew our hand in the first place, the 52 choose 5.
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Think about if they are all going to be spades.
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If they are all going to be spades then we are really saying it is the number of ways
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to choose 5 spade out of how many spades are there total?
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There are 13, the ace through the king of spades, out of 13.
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According to our logic before that is exactly 13 choose 5.
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If we wanted to express that in factorials, that is 13! ÷ 5! ÷ 13 -5! which is 13! ÷ 5! × 8!.
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That is our number of ways to choose 5 spades out of, there are only 13 in the deck.
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That is the number of ways that we can get 5 spades.
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Our probability of A is the number of outcomes in A ÷ the number of outcomes in S.
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We already calculated both of those, it is 13 choose 5 ÷ the total number of poker hands we can get is 52 choose 5.
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We could simplify that down if we wanted.
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13! ÷ 5! × 8!.
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We already saw above that 52 choose 5 is 52! ÷ 5! × 47!.
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We could simplify that a bit further.
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We can cancel the 5! and flip up the denominator.
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13! ÷ 52!.
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13! ÷ 8! × 52!.
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The 47 would flip up to the numerator and the 5! would cancel.
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I'm going to make no attempt to try to actually calculate that as a number.
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It would be a very small number.
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But that is our probability, if you draw 5 cards out of the deck that you will get all 5 cards will be spades.
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That is of the end of our 6th example there.
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Let me recap just quickly what we did on the second slide here.
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Our event was the number ways to get a poker hand of all spades.
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Using our strategy from before that is 13 choose 5, remember we have learned that notation on the previous slide.
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If you do not believe that, another way to think about that is to say how many ways can we get all 5 spades.
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If you are choosing 5 spades, there are 13 ways to get the first one.
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Once you have picked one spade, there are 12 ways after that to get the second one.
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11 ways once you picked the first two, × 10 × 9.
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There is your 5 spades but that over counts because the 5 spades could have come in any order.
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The number of possible orders they could arrive in is 5 × 4 × 3 × 2 × 1.
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That simplifies, we have 5! in the denominator and the numerator we can cleverly write it as 13! ÷ 8!.
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Because we are multiplying all the numbers down from 13 but we are cutting off all the ones down from 8.
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We can get that as 13! ÷ 8!.
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That is another way to get to that same answer there for the event that we are interested in.
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To find the probability, we just take the number of ways to get to our event ÷
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the total number of outcomes 13 choose 5 ÷ 52 choose 5.
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We get a kind of cumbersome expression involving factorials in our answer.
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That wraps up our last example and that wraps up our introductory lecture here on www.educator.com, for probability.
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We went over some terminology, we practiced some basic probability using counting techniques.
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On the next set of lectures, we will develop these counting techniques further
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and we will have complicated probability problems.
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I hope you will stick around for that.
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These are the probability lectures here on www.educator.com and my name is Will Murray.
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Thanks, bye.