Exam Questions – Binomial distribution

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What is the probability that the first strike comes on the third well drilled? Note that X is technically a geometric binomial distribution example questions variable, since we are only looking for one success. Since a geometric random variable is just a special case of a negative binomial distribution example questions random variable, we'll try finding the probability using the negative binomial p.

It is at the second equal sign that you can see how the general negative binomial problem reduces to a geometric random variable problem. What is the mean and variance of the number of wells that must be drilled if the oil company wants to set up three producing binomial distribution example questions Eberly College of Science. Geometric and Negative Binomial Distributions.

The mean number of wells is: Introduction to Probability Section 2: Discrete Distributions Lesson 7: Discrete Random Variables Lesson 8: Mathematical Expectation Lesson 9: Moment Generating Functions Lesson The Binomial Distribution Lesson The Poisson Distribution Section 3: Continuous Distributions Section 4: Bivariate Distributions Section 5: Distributions of Functions of Random Variables. Hypothesis Testing Section 8: Nonparametric Methods Section 9: Bayesian Methods Section

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Where the Bernoulli Distribution asks the question of "Will this single event succeed? The relation between the Bernoulli and binomial distributions is intuitive: The binomial distribution is composed of multiple Bernoulli trials.

This number of successes is represented by the random variable X. For a refresher on factorials n! Let's walk through a simple example of the binomial distribution. We're going to use some pretty small numbers because factorials can be hard to compute. We are going to ask five random people if they believe there is life on other planets.

We want to ask the question: We know all the values that we need to plug into the equation. This yields the equation:. Here are the probabilities for all the possible values of X. What can we learn from these results? Well, first of all we'll see that it's just a little more likely that only one person will confess to believing in life on other planets.

So the probability that any two people both believe in extraterrestrial life is 0. Now, for two out of five people to believe in extraterrestrial life, two conditions must be satisfied: The probability of two out of five people believing in extraterrestrial life would thus appear to be 0. However, in doing this, we are only considering the case whereby the first two selected people are believers.

How do we consider cases such as that in which the third and fifth people are believers, which would also mean a total of two believers out of five? The answer lies in combinatorics. Bearing in mind that the probability that the first two out of five people believe in extraterrestrial life is 0. This is why we multiply by C n,k. The probability of having any two of the five people be believers is ten times 0.

We can now rewrite the summation as. We now see that the summation is the sum over the complete pmf of a binomial random variable distributed Bin m, p. This is equal to 1 and can be easily verified using the Binomial theorem. We have already calculated E[ X ] above, so now we will calculate E[ X 2 ] and then return to this variance formula:. We can use our experience gained above in deriving the mean. We use the same definitions of m and w. The first sum is identical in form to the one we calculated in the Mean above.

It sums to mp. The second sum is 1. From Wikibooks, open books for an open world. Statistics Introduction What Is Statistics?

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