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# Theoretical probability distribution example: multiplication

AP.STATS:
VAR‑5 (EU)
,
VAR‑5.A (LO)
,
VAR‑5.A.1 (EK)
,
VAR‑5.A.2 (EK)
,
VAR‑5.A.3 (EK)
CCSS.Math:

## Video transcript

we're told that kai goes to a restaurant that advertises a promotion saying one in five customers get a free dessert suppose kai goes to the restaurant twice in a given week and each time he has a one-fifth probability of getting a free dessert let x represent the number of free desserts chi gets in his two trips construct the theoretical probability distribution of x alright so pause this video and see if you can work through this before we do it together all right so first let's just think about the possible values that x could take on this is the number of free desserts he gets and he visits twice so there's some world in which he doesn't get any free desserts so that's zero in his two visits maybe on one of the visits he gets a dessert and the other one he doesn't and maybe in both of his visits he actually is able to get a free dessert so he's going to have some place from 0 to 2 free desserts in a given week so we just have to figure out the probability of each of these so let's first of all think about the probability let me write it over here the probability that capital x is equal to zero is going to be equal to what well that's going to be the probability that he doesn't get a dessert on both days and it's important to realize that these are independent events it's not like the restaurant's gonna say oh if you didn't get a dessert on one day you're more likely to get another day or somehow if you got it on a previous day you're less likely on another day that they are independent events so the probability of not getting it on any one day is four out of five the probability of not getting it on two of the days i would just multiply them because they are independent events so 4 over 5 times 4 over 5. so the probability that x is equal to 0 is going to be 16 twenty-fifths sixteen over twenty-five now what about the probability that x is equal to one what is this going to be well there are two scenarios over here there's one scenario where let's say on day one he does not get the dessert and on day two he does get the dessert but then of course there's the other scenario where on day one he gets the dessert and then on day two he doesn't get the dessert these are the two scenarios where he's going to get x equals one and so if we add these together let's see four-fifths times one-fifth this is going to be four over twenty-five and then this is going to be four over twenty-five again and you add these two together you're going to get eight twenty-fifths and then last but not least and actually we could figure out this last one by subtracting 16 and 8 from 25 which would actually give us 1 25th but let's just write this out the probability that x equals 2 this is the probability gets a desert on both days so one fifth chance on day one and one fifth chance on the second day so one-fifth times one-fifth is 1 25th and you can do a reality check here these all need to add up to one and they do indeed add up to 1. 16 plus 8 plus 1 is 25 so 25 25 is what they all add up to and we're done