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## Theoretical & empirical probability distributions

Current time:0:00Total duration:3:25

# 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

- [Instructor] We're told
that Kai goes to a restaurant that advertises a promotion saying, "1 in 5 customers get a free dessert!" Suppose Kai goes to the
restaurant twice in a given week, and each time he has a 1/5 probability of getting a free dessert. Let X represent the
number of free desserts Kai gets in his two trips. Construct the theoretical
probability distribution of X. All right, 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 0 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 someplace 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 0 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 the other day," or, "Somehow if you got
it on a previous day, you're less likely than another day." That they are independent events. So the probability of not
getting it on any one day is four out of five, and the probability of not
getting it on two of the days, I would just multiply them because they are independent events. So four over five times four over five. So the probability that X is equal to 0 is going to be 16/25, 16 over 25. Now what about the probability
that X is equal to 1? 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 1. And so if we add these
together, let's see. 4/5 times 1/5, this is going to be 4/25, and then this is going to be 4/25 again. And you add these two together, you're going to get 8/25. 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 that
he gets a dessert on both days. So 1/5 chance on day one and 1/5 chance on the second day. So 1/5 times 1/5 is 1/25. And you can do a reality check here. These all need to add up to one. And they do indeed add up to one. 16 plus 8 plus 1 is 25. So 25/25 is what they all add up to. And we're done.