Cumulative geometric probability (less than a value) | AP Statistics | Khan Academy

Lilliana runs a cake decorating business for which 10% of her orders come over the telephone. Let's see ( C ), the number of cake orders Lilliana receives in a month until she first gets an order over the telephone.

Assumed a method of placing each cake order is independent, so if we assume a few things as a classic geometric random variable, what tells us that? Well, a giveaway is that we're gonna keep doing these independent trials where the probability of success is constant. There's a clear success—a telephone order in this case is a success. The probability is 10% of it happening, and we're gonna keep doing it until we get a success. So, classic geometric random variable.

Now they asked us to find the probability—the probability that it takes fewer than five orders for Lilliana to get her first telephone order of the month. So, it's really the probability that ( C < 5 ).

So, like always, pause this video and have a go at it. Even if you struggle with it, that's even better. Your brain will be more primed for the actual solution that we can go through together.

All right, so I'm assuming you've had a go at it. There's a couple of ways to approach it. You could say, well, look, this is just going to be the probability that ( C = 1 ) plus the probability that ( C = 2 ) plus the probability that ( C = 3 ) plus the probability that ( C = 4 ), and we can calculate it this way.

What is the probability that ( C = 1 )? Well, the probability that her very first order is a telephone order is ( 0.1 ).

What's the probability that ( C = 2 )? Well, the probability that her first order is not a telephone order is ( 1 - 0.1 ), so there's a 90% chance it's not a telephone order, and that her second order is a telephone order.

What about the probability ( C = 3 )? Well, her first two orders would not be telephone orders and her third order would be.

Then ( C = 4 ): well, her first three orders would not be telephone orders, and her fourth one would.

We could get a calculator maybe and add all of these things up, and we would actually get the answer, but you probably wonder, well, this is kind of hairy to type into a calculator; maybe there is an easier way to tackle this, and indeed there is.

So think about it: the probability that ( C < 5 ) is the same thing as ( 1 - ) the probability that we don't have a telephone order in the first four. So ( 1 - ) the probability that no telephone order in first four orders.

So what's this? Well, because this is just saying we, you know, what's the problem we do have an order in the first four? So it's the same thing as ( 1 - ) the probability that we don't have an order in the first four.

This is pretty straightforward to calculate. So this is going to be equal to ( 1 - ) and let me do this in another color so we know what I'm referring to.

So what's the probability that we have no telephone orders in the first four orders? Well, the probability on a given order that you don't have a telephone order is ( 0.9 ), and then if that has to be true for the first four, well, it's going to be ( 0.9 \times 0.9 \times 0.9 \times 0.9 ) or ( 0.9^4 ).

So this is a lot easier to calculate. So let's do that. Let's get a calculator out.

All right, so let me just take ( 0.9^4 ) which is equal to—and then let me subtract that from one, so let me make that negative and then let me add one to it—and we get, there you go, ( 0.3439 ).

So this is equal to ( 0.3439 ), and we're done. That's the probability that it takes fewer than five orders for her to get her first telephone order of the month.