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# Geometric random variables introduction

Distinguishing between geometric and binomial random variables.

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• 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.
• 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.
• 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!

• The captions managed to incorrectly spell 'googol' lol
• 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?
• 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
• 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?
• 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)