Conditional Probability I

George Ingersoll
A free video tutorial from George Ingersoll
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Lecture description

Introduction to conditional probability and how to solve using the fundamental probability equation.

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Workshop in Probability and Statistics

This workshop will teach you the fundamentals of statistics in order to give you a leg up at work or in school.

21:37:01 of on-demand video • Updated April 2020

  • By the end of this workshop you should be able to pass any introductory statistics course
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English Have folks in this video we're going to talk about conditional probability and unfortunately conditional probability is one of those concepts that trips up a lot of people in the early going with probability and statistics. Let's do that because it doesn't have to be at its fundamental core. Conditional Probability is not so much of a departure from the fundamental probability you've studied thus far and I say that because with conditional probability you can continue to use this equation here . Number of outcomes you're looking for over number of total possible outcomes. However with conditional probability both the numerator and denominator are going to change. Let's take a look at an example to illustrate what conditional probability is. So those of you who watched my previous videos can we've seen examples along these lines before. We have in this example a single drop from a deck of 52 cards. Is taking one card out of a deck of 52 cards. Two events in a draw heart in that single draw and Event B is the office card and that single drop. So just take on their own. We know the probability of event. Hey let's go ahead and use actual 10 probability of event a happening is going Matz's equation number of tackles we're looking for a number of total possible outcomes. There are 13 parts in the deck total possible outcomes are 52 52 cards. Probability of event be happening. There are 12 face cards in the deck. And of course there are still 52 cards. So here's where conditional probability comes in. Suppose I said to you that the card that you're drawing right now is a heart. I checked already. It's definitely hard. That's the condition you're going to draw hard. But now what is the probability that you're going to draw a face card. So you know it's a hard. What's the probability that you're going to draw a face card. Let's think about this from a commonsense perspective before throwing a question at you because you can do this. A lot of these conditional probability problems was common sense. Well I've told you the heart. There are 13 hearts in the deck right. So let's start there. If there are 13 hearts that's the number of total possible outcomes. That's this here. I you can only draw one of those 13 hearts because I told you with a heart and how many cards are there among the hearts. Well there's the Jack of Hearts. Queen of Hearts King. There's three so that the probability of drawing a face card. If you know you're drawing a heart is going to be 3 out of 13. And the way I would write this is probability of the probability of drawing a face card given a meaning . The condition is in that we know you're drawing a hard and you separate the two of them. Essentially we read this this way. Often times is a and so same condition. He will say given as in this is a given that a is happening. So you read this equate this. You know what I just wrote here as probability of be drawing a face card. Given that the card you are drawing is a heart given that event A is A. It's a given. It's definitely happening. So proud. P be given a s to say it's probably getting a face card. If you know you're getting heart all right let's flip this in and use an example. The other way around. Let's say I told you it's a face card that that draw is going to be a face card. What is the probability that you've drawn heart. OK so this as you can probably deduce is going to be p. I was proud of getting at heart. If you know you've got a face card p b c p a given B. All right well there are 12 face cards in the deck right there is the jack queen king of clubs on his hard spaces three times for twelfth face cards in the deck or total possible outcomes or 12 and among the face cards. How many of them are hearts still three. So our probability of a given problem is getting harder. Given we've got a face card is going to be 3. 12. Now there is an equation for this and I want you to write it down because it's important. Let's do this one down here w a B. This can also be written as probability of the intersection of A and B over the probability of B this is the equation to figure out conditional probability. So write this down. You can flip this all over. You know do the exact same thing here. Probability of the intersection which is probability of a intersection B and probably the intersection and are the exact same thing. You don't have to write them. It doesn't matter how you write it. So this is going to be the same regardless. But the denominator here is going to be probability. OK let's just do a sanity check to make sure this thing works. So for probability of a start up here just for that I don't know just because it's it's higher up on the page. Probability of be given. I think that what is the probability of B intersection. OK. OK. That's going to be the intersection of A and B and if you don't know this go back to the previous videos . Probability of the start of the intersection of A B is the heart face cards right. The cards that are both hearts and face card so you're just drawing once you're saying the intersection of A and B is the jackpot at the Queen of Hearts and the King of Hearts. That's the probability of getting that on a single draw is three out of 52 three total possibilities out of a deck of 52 cards right. And probability of A which we already defined here is 13 out of 52. So we could just cancel out these 52 as in this equation and we get three other team. Hey worked great. Same thing down here. I just told you probability of the intersection of A B is the same things. Probably probably the same thing. The intersection of B and they. So that's going to be three out of 52 and the probability of event B is there's 12 face cards in a deck of 52 cards. You can tell this is working people three twelfths which is exactly what we got previously. So before we move on the main thing I want to point out about this example is despite the fact that we learned a new equation here it really all comes back to this number of outcomes we're looking for the total possible outcomes but with conditional probability you've reduced both the numerator and denominator . Like I said we would for instance for probability of the drawing a face card when we know that that card is a heart. That was the first example we looked at if we know the cards heart probably drawing a face card. We come back to here and say instead of 52 cards now we're dealing with just the hearts just 13 cards and the numerator is also reduced We're not talking about all the face cards in the deck. We're now just talking about the face cards that are hard. So that's the jackpot with the queen part the king part. So just three over 13. Now I want to also talk about independent events. And if you'll recall from the previous video when we talked about independent events we said independent events are events that do not influence each other meaning event A does not influence event B and vice versa. They have no effect on each other's probabilities. The example I gave you I believe had to do with rolling two dice. Say red dye and a white tie whatever you roll on the white tie is not going to influence whatever you get on the red dye and vice versa. They have nothing to do with each other. Well when independent events it has invented events independent events are different with conditional probability. And when we say worked out by independent events we can use this equation. Let me just walk you through it probably a given B is equal to probability of A and B are independent . Let's think about this from intuitive standpoint independent of us and B are independent. They have no influence on each other. So if I tell you it's a condition that proud that that event B is met does that affect probability of event. A No. So if I say Event B is the condition it makes no difference because they're independent they have no influence on each other. So the probability of a given B is just going to be the same thing as probability. Let's keep this in mind and re-examine our previous example here. Here we have a probability of let's let's take this first example. Probability of drawing a face card if we know the heart of the heart. Are the events A and B independent. Let's let's use this equation here. If they are independent then a given B is going to be equal to probability. Now in this case reading be given AA is the same thing. But what we should be looking at is what we should be examining here is is the probability of drawing a face card B given that we know the card is a heart get the same probability as just drawing based hard. Let's look at it. Well we just can't play this out using this equation here. That was three over 13. What is the probability of just drawing a base card. Well we calculate that earlier. It's 12 over 52. Now it's not exactly the same fraction but they are there. This is just a reduced form of this. You multiply both in a random moment by four you get 12 or 52. He says that works. And just as a sanity check. Let's go and check this one out as well. The probability of if these if A and B are independent which we suspect that they are at this point probability of drawing a heart if we know that the card has a face card should be the same as the probability of just Rheinhardt should be equal to probability. Well probability of drawing hard you know that the car is the face card and calculated that I use in the equation we got 3 over 12 which is also well known to be one fourth and the probability of drawing the heart P-A is also one fourth It's just a different version of it. 1252 is also one fourth. So hey looks like the events A and B are independent. Now let's look at an example that we would not have independence. Ok so on screen I've put a new scenario. It's actually it's similar to a previous scenario that we looked at. It's the doctor full of socks when we walk through we got eight pairs of black socks. But among those there's actually six pairs of black dress socks and two players of black sweat socks . We also have five total blue socks but there's three pairs of dressage. Blue dress socks two pairs of blue sweat socks. We also have seven pairs of white socks. But they're also socks. So what does that tell us. What else is a few things. We've got the black socks eight pairs of black socks. Yes I know. We have five pairs of socks and seven pairs white socks that gives us actually 20 total socks pairs of socks 20 pairs of socks total. Also looks like we have 6 3 6 3 9 pairs dress socks and 11 pairs of sweat socks. I get some good information here and I'm going to give it to events we have. We're getting these scenarios. We're going to grab one pair of socks that looks like it says Saks. But you know I'm talking about one pair of socks and event A is going to be you grab black socks and Event B is dress socks. Now if this were my sock drawer this scenario wouldn't even be possible because I don't ball up my socks into pairs. All just hanging around. So this would be a complete mess. I want to know. I've I've already given you the answer. But I want to know are these events independent. Are A and B independent getting black socks and getting dress socks. I'm telling you now they're not. But I want I want to prove it. So you mean at this point I want to pause the video and see if you can prove yourself or if you're not quite there yet then we'll do it together. Who's ready for my answer. The question is based on this equation here. If they are independent then the probability of a given B is going to be equal to the probability of A. All right. So the probability is let's write this down a kid and B it's going to be equal to the probability. OK. So first of all what is the probability of a given B. B is there dress socks that's been given to us and say is there black socks. So we're saying if we know these are dress socks What is the probability that they're black. All right. Now we have total nine pairs of dress socks six of them are black and three of them are blue. So let's think about this conditional probability here we're just thinking of possible outcomes we're looking for over total possible outcome. Well we can do this. So this is a question mark. This is Bretons right now we're saying is it are these equal if they are then we have independence of these events and we're saying probably a given B their dress socks. We know there's only 11 or nine pairs of dress socks total. Those are the outcomes the total possible outcome is only 9 percent rest because we've said they're definitely dress socks that's on our condition. Now one knows probably they're black. Well the dress that we're made up of six pairs of black socks and three pairs of socks. So either it's black. If we know their dress aren't we just going to be six over nine now. What's the probability of hey I believe age is just that. Probably we have black socks from our our drawer pick. Well there's the president black socks and 20 pairs total in the softer. So that could be eight over 20. So the question is are these two events independent. Well this is they would be independent if this were true but it is clearly not true. Six over nine doesn't equal even 20 so events a and b are not independent.