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### Course: AP®︎/College Statistics > Unit 8

Lesson 7: The geometric distribution- Geometric random variables introduction
- Binomial vs. geometric random variables
- Geometric distribution mean and standard deviation
- Geometric distributions
- Probability for a geometric random variable
- Geometric probability
- Cumulative geometric probability (greater than a value)
- Cumulative geometric probability (less than a value)
- TI-84 geometpdf and geometcdf functions
- Cumulative geometric probability
- Proof of expected value of geometric random variable

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# Geometric random variables introduction

Distinguishing between geometric and binomial random variables.

## Want to join the conversation?

- Is there an example of a case where:

1. Each trial has the same probability AND

2. Trial results are NOT independent?

I can see that there can be a case where trial results are independent with different probability, but I am just curious if the reverse is also true.(6 votes)- Sure. Consider an experiment in which the success of the second trial is determined by the success of the first. That is, if the first is success then the second is success and if the first is failure then the second is failure. Each trial would have the same probability of success and yet their results would not be independent.(4 votes)

- When will you make a geometric mean video??(1 vote)
- The geometric mean of a list of n non-negative numbers is the nth root of their product.

For example, the geometric mean of the list 5, 8, 25 is cuberoot(5*8*25) = cuberoot(1000) = 10.

It has been proven that, for any finite list of one or more non-negative numbers, the geometric mean is always less than or equal to the (usual) arithmetic mean, with equality occurring if and only if all the numbers in the list are the same.

Have a blessed, wonderful day!(5 votes)

- 3:50

The captions managed to incorrectly spell 'googol' lol(2 votes) - Is it still a geometric random variable if you ask yourself how many trials will it take till you make two successes in a row?(2 votes)
- When you're looking for two successes in a row, the distribution is no longer geometric. The geometric distribution measures the number of trials needed to get the first success. For scenarios involving "two successes in a row," you're entering the realm of a different kind of distribution, often explored through Markov chains or other statistical models that account for specific patterns in sequences of trials.(1 vote)

- Thank you Khan Academy for the details in this video. From Viet Nam with love(2 votes)
- The video claims Y is not a binomial random variable because we can't say how many trials it might take to roll a 6. But realistically, if we roll the die, say, 80 times, the probability that we won't roll a 6 is only 0.000046%, which in any answer to a Khan Academy quiz question would be rounded to 0%. So, as long as we're willing to accept a very very small amount of uncertainty (and we all do, in every action we take), then we could "convert" this geometric random variable into a binomial random variable easily, by choosing some large number of times we are going to roll it. And in the video's example we can confidently handle Y as a binomial random variable... right?(2 votes)
- Theoretically, setting a large number of trials could approximate the conditions for a binomial random variable because you cap the trials, making them finite. However, this adaptation fundamentally changes the variable from measuring "the number of trials until success" to "the number of successes in a set number of trials," which are different concepts. The essence of a geometric distribution is the uncertainty in the number of trials until the first success, which is lost in the binomial approximation. Although practically, for a very large number of trials, the difference in interpretation might seem subtle, the underlying principles distinguishing binomial and geometric distributions remain significant.(1 vote)

- Why are the trial results independent for Y? It is "until I get a 6" so if for example I get a 6 in my first trial then I won't need to roll the dice again(1 vote)
- The independence of trials for variable Y is because the outcome of each roll of the die does not affect the outcome of subsequent rolls. Even if a 6 is rolled on the first trial, which would stop further trials, each trial's outcome is still independent of the others. Independence here refers to the statistical independence of events, not the procedural cessation of trials after a success.(1 vote)

- Can anyone explain why the answer and hint makes sense?

It seems the question hasn't mention anything about selection without replacement to me

Question: A class with 25 students randomly selects a different student each week to bring a class snack. Of the students, 8 percent have food allergies. Let T be the number of weeks until a student with a food allergy is selected to bring the snack.

Answer:T is neither type of variable.

Hint: In this situation, the students are selected without replacement from a small population, so each trial is not independent.(1 vote)- In the scenario of selecting a student to bring a snack, the trials are not independent because the population of students decreases with each selection (no replacement), altering the probability of selecting a student with a food allergy in subsequent weeks. This violates the independence criterion for both binomial and geometric distributions, making variable T neither type of random variable.(1 vote)

- for variable Y - the one with e # of rolls to get a 6. Would it be a binomial random variable if I change the variable to this:

1 if 6 was rolled in the 1st roll

2 otherwise

based on Sal's argument I would would say yes, because I eliminated the issue of non-fixed number of trials. It will be always 1, I would never need to try a second roll.(1 vote)- If you change the definition of the variable such that it takes a value of 1 if a 6 is rolled on the first try and 2 otherwise, then yes, it becomes a binomial random variable. This is because you have clearly defined a fixed number of trials (1 trial), and the outcome of that trial can be categorized as a success or failure, fulfilling the requirements for a binomial distribution. The experiment now has a deterministic number of trials (exactly one), with two possible outcomes.(1 vote)

- Similarly to the one provided for the binomial random variable, what's the checklist for the geometric random varialble?(1 vote)
- There must be a sequence of independent trials.

Each trial has two possible outcomes (success or failure).

The probability of success, p, remains constant for each trial.

The variable of interest is the number of trials until the first success occurs.(1 vote)

## Video transcript

- [Narrator] So I have two,
different random variables here. And what I wanna do is think about what type of random variables they are. So this first random variable, x, is equal to the number of sixes after 12 rolls of a fair die. Well this looks pretty much like a binomial random variable. In fact, I'm pretty confident it is a binomial random variable and we can just go down the checklist. The outcome of each trial can be a success or failure. So, trial outcome success or failure. It's either gonna go either way. The result of each trial is independent from the other one. Whether I get a six on the third trial is independent of whether I got a six on the first or the second trial. So result, let me write this, trial, I'll just do a shorthand trial, results independent, independent, that's an important condition. Let's see, there are a
fixed number of trials. Fixed number of trials. In this case we're gonna have 12 trials. And then the last one is, we have the same probability on each trial. Same probability of success probability on each trial. So yes indeed, this met
all of the conditions for being a binomial,
binomial random variable. And this was all just a little bit of review about things that we have talked about in other videos. But what about this thing
in the salmon color? The random variable y. So this says the number of rolls until we get a six on a fair die. So this one strikes us as
a little bit different. But let's see where it
is actually different. So, does it meet that the trial outcomes that there's a clear success
or failure for each trial? Well yeah, we're just gonna keep rolling. So each time we roll, it's a trial. And success is when we get a six. Failure is when we don't get a six. So the outcome of each trial can be classified as either
a success or a failure. So it meets, maybe I'll put the checks right over here, it meets
this first constraint. Are the results of each trial independent? Well whether I get a six on the first roll or the second roll, or the third roll, or the fourth roll, or the third roll, the probabilities shouldn't be dependent on whether I did or didn't
get a six on a previous roll. So, we have the independents. And we also have the same probability of success on each trial. In every case it's a 1/6 probability that I get a six, so this stays constant. And I skipped this third
condition for a reason. Because we clearly don't have
a fixed number of trials. Over here we could roll 50
times until we get a six. The probability that we'd have to roll 50 times is very low. But we might have to roll 500 times in order to get a six. In fact, think about what
the minimum value of y is and what the maximum value of y is. So the minimum value that
this random variable can take, I'll just call it min y, is equal to what? Well, it's gonna take at least one roll. So that's the minimum value. But what is the maximum value for y? And I'll let you think about that. I'll assumed you thought about
it, if you paused the video. Well, there is no max value. You can't say, "Oh it's a billion." Because there's some probability that it might take a
billion and one rolls. It is a very, very, very,
very, very, very small probability, but there's some probability. It could take a Google
rolls, a Google plex rolls. So you can imagine where this is going. So this type of random variable, where it meets a lot of the constraints of a binomial random variable. Each trial has a clear
success or failure outcome. The probability of success
on each trial is constant. The trial results are
independent of each other. But we don't have a
fixed number of trials. In fact, it's a situation, we're saying, "How many trials do we need to get, "to we need to have until we get success?" Maybe that's a general way of framing this type of random variable. How many trials until success? While the binomial random variable was, how many trials, or how many successes, I should say, how many successes in finite number of trials? So if you see this general form and it meets these conditions, you can feel good it's a binomial random variable. But if we're meeting this condition, clear success or failure outcome, independent trials, constant probability, but we're not talking about the successes in a finite number of trials. We're talking about how
many trials until success? Then this type of random variable is called a geometric random variable. And we will see why, in future videos it is called geometric. Because the math that involves the probabilities of various outcomes looks a lot like geometric growth, or geometric sequences and series that we look at in other
types of mathematics. And in case I forgot to mention, the reason why they're called binomial random variables is because when you think about the probabilities
of different outcomes, you have these things called
binomial coefficients, based on combinatorics. And those come out of things like Pascal's Triangle and when you take a binomial to ever increasing powers. So that's where those words come from. But in the next few videos, the important thing is to recognize the
difference between the two. And then we're gonna start thinking about how do we deal with
geometric random variables.