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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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# Probability for a geometric random variable

Finding the probability for a single outcome of a geometric random variable.

## Want to join the conversation?

- This is the second video (as Feb 2019) in the Geometric Variables playlist (learning module). It asks us to "pause the video and have a go at it" but it hasn't introduced the method for answering questions with geometric random variables yet. (It wasn't introduced in the introductory video). It seems that the difference between binomial and geometric, in setting up the problem, is that binomial problems involve an "n choose k" (nCk) before multiplying probability of individual fails or success. With geometric, I'm guess the "nCk" can be avoided? If so, this important distinction should be emphasized in the video.(9 votes)
- I had the same thought when he said we could have a go at it even though the concept is not yet introduced. But I chose to have a go at it anyway and I could solve it pretty easily. It was fairly simple.

Just like we find the probability for binomial variables, that is all we have to do except we don't need the combinations here because the question provides us with it (3rd throw, so MissMissMake). There is only one arrangement here unlike the questions for binomial variables where it asked "probability of making exactly 3 out of 5 throws".(1 vote)

- I guess this problem can be solved using Negative binomial distribution(as it is the generalized version) ?(2 votes)
- Yes, you can solve this problem using the negative binomial distribution, which is a generalization of the geometric distribution. In the negative binomial distribution, you're interested in the number of trials needed to achieve a predetermined number of successes, whereas in the geometric distribution, you're interested in the number of trials until the first success. Since the question asks for the probability of the first success occurring on the third attempt, which is a specific instance of the geometric distribution, you can directly use the properties of the geometric distribution to solve it.(1 vote)

- how to interpret the expected value of 4 in this case (1/p=1/.25=4) Isn't that the shots required to make his first hit? so how can we interpret this 3?(1 vote)
- The expected value of a geometric random variable, E[X], where X represents the number of trials until the first success, is calculated as 1/p, where p is the probability of success. In this case, p = 0.25, so E[X] = 1/0.25 = 4. This means that on average, Jeremiah is expected to make his first successful three-point shot after 4 attempts. However, the specific instance of M = 3 refers to the probability of the first successful shot occurring on the third attempt, which is a different concept from the expected value. The expected value gives us the average number of trials until the first success, while M = 3 represents a single outcome where the first success happens specifically on the third attempt.(1 vote)

- How would one calculate the number of trials until the first success?(1 vote)
- To calculate the number of trials until the first success in a geometric random variable scenario, you can use the formula p(k) = (1 − p)^k−1 × p, where p is the probability of success and k is the number of trials until the first success. In this case, p is the probability that Jeremiah makes a three-point shot, which is 0.25, and you're asked to find the probability that M = 3, meaning the first successful shot occurs on the third attempt.(1 vote)

- Where did the 1/4 come from(0 votes)
- 25% or 0.25 = 1/4(2 votes)

- I don’t understand

I don’t see the answer(0 votes)- Let's see.

His chance of to miss his shot = 3/4.

His chance of to make his shot = 1/4.

His probability of miss his first two shots and make his third will be = 3/4*3/4*1/4 = 9/64, in decimals 0,140625 or 14%. As you prefer.

I'm not quite fluent in english, so if I wrote something wrong, please correct me, that's it.(4 votes)

- I think probability for geometric random variable helps solve the problems easily?(0 votes)
- Indeed, the probability for a geometric random variable helps solve problems involving the number of trials until the first success easily. The geometric distribution captures the probability of achieving the first success on a specific trial, given a constant probability of success on each trial. In this case, the probability of Jeremiah making a three-point shot on any given attempt is 0.25, allowing us to calculate the probability of his first successful shot occurring on the third attempt.(1 vote)

- A bag with 10 marbles is shown, A marble was chosen from the bag at random. What probability that it is black. Write answer as a fraction in simplest form.(0 votes)
- Not enough information. You have to know how many black marbles are in the bag to solve this problem. For example 1 black marble, P = 1/10 , 2 black marbles, P = 2/10 = 1/5, etc.(1 vote)

## Video transcript

- [Instructor] Jeremiah makes 25% of the three-point shots he attempts. Far better than my percentage. For warm up, Jeremiah likes to shoot three-point shots until
he successfully makes one. Alright, this is the telltale signs of a geometric random variables. How many trials do I have to take until I get a success? Let M be the number of
shots it takes Jeremiah to successfully make his
first three-point shot. Okay, so they're defining
the random variable here, the number of shots it takes, the number of trials it takes until we get a successful
three-point shot. Assume that the results of
each shot are independent. Alright, the probability
that he makes a given shot is not dependent on
whether he made or missed the previous shots. Find the probability that
Jeremiah's first successful shot occurs on his third attempt. So, like always, pause this video and see if you can have a go at it. Alright, now let's work
through this together. So, we wanna find the probability. So, M is the number of shots it takes until Jeremiah makes
his first successful one and so, what they're really asking us, find the probability
that M is equal to three, that his first successful shot occurs on his third attempt. So, M is equal to three. So, that the number of
shots it takes Jeremiah to make his first
successful shot is three. So, how do we do this? Well, what's the probability
of that happening? Well, that means he has to
miss his first two shots and then make his third shot. So, what's the probability of
him missing his first shot? Well, if he has a 1/4 chance of making his shots, he
has a 3/4 probability of missing his shots, so this will be 3/4, so he misses the first shot, times, he has to miss the second shot and then he has to make his third shot, so there you have it. That's the probability. Miss, miss, make and so, what is this going to be? This is equal to nine over 64ths. So, there you have. If you wanted to have this as a decimal, we could get a calculator out real fast, so this is nine divided by 64 is equal to roughly 0.14. Approximately 0.14 or another way to think about it is a roughly 14% chance or 14% probability that his first successful shot occurs in his third attempt.