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Theoretical probability distribution example: multiplication | Probability & combinatorics

3m read
·Nov 10, 2024

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 he 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.

Alright, 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." 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 twenty-fifth. But let's just write this out.

The probability that X equals 2 is the probability he gets a dessert 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 twenty-fifth.

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 twenty-fifths is what they all add up to. And we're done.

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