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For more information, please see full course syllabus of Probability
For more information, please see full course syllabus of Probability
Probability Negative Binomial Distribution
Lecture Description
In this lesson, we are going to talk about the negative binomial distribution. The negative binomial distribution describes a sequence of trials, each of which can have two outcomes, success or failure. In the binomial distribution, we decide ahead of time how many times we are going to flip a coin and then we keep track of the number of heads after it is all over. With the negative binomial distribution, we decide ahead of time how many heads we want to see and we flip for as long as it takes to get that number of heads. You will learn the formula for the negative binomial distribution and you'll see some of its key properties.
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3 answers
Mon Mar 9, 2015 9:26 PM
Post by Anhtuan Tran on February 26, 2015
Hi Dr Murray,
I have another question on example 4, part b.
Let X be the numbers of the interviews.
We're trying to calculate P(X>= 10). P(X>=10) also means that we don't get 3 qualified applicants in the first 9 interview and then it becomes a simple binomial distribution problem. I totally agree with you about that.
But here is my other approach. Assume that I want to find P(X <= 9) and then my P(X>=10) = 1 - P(X<=9).
P(X<=9) means that as long as I get the 3 qualified applicants in the first 9 interviews. Therefore, P(X<=9) = 9C3 . p^3 . q^6. And then I just subtract it from 1.
However, I didn't get the same answer. I have a feeling that something is wrong with P(X<=9), but I couldn't figure out the reason why it didn't work.
Thank you.
3 answers
Fri Feb 27, 2015 1:19 PM
Post by Anhtuan Tran on February 21, 2015
Hi Professor Murray,
I was trying to figure out what kind of distribution of this problem was, but I couldn't. Here is the problem: There are 2n balls with n different colors. Each color has 2 balls. Draw balls until getting a ball with the color that has appeared before. Find the distribution.
There are two cases:
i/ drawing with replacement
ii/ drawing without replacement.
How do you approach this problem?
Thank you.
4 answers
Mon Jan 12, 2015 10:24 AM
Post by Anton Sie on December 26, 2014
Hi, I have a question about example I: how do you calculate the same chance, but without replacement?