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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.
• 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) ?
• 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
• 25% or 0.25 = 1/4
• I don’t understand
• 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.