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Question:

(a) What does it mean to say that a random variable has a geometric distribution with parameter \(p\) ?

b) What is the mean of a geometric distribution with parameter \(p\) ?

Short Answer

Expert verified

Answer

(a) The probability distribution of the first success among independent trials with a constant probability \(p\) of success is the resultant answer.

(b) The resultant answer is\(\frac{1}{p}\).

Step by step solution

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01

Given data

The given data is that a random variable has a geometric distribution.

02

Concept of Geometric probability

Geometric probability: \(P(X = k) = {q^{k - 1}}p = {(1 - p)^{k - 1}}p\).

03

Simplify the expression 

(a)

A random variable that has a geometric distribution with probability \(p\) of success represent the probability distribution of the first success among independent trials with a constant probability \(p\) of success.

Definition geometric probability: \(P(X = k) = {q^{k - 1}}p = {(1 - p)^{k - 1}}p\) that is, the probability of the first success occurring on the \(k - th\) trial is \({(1 - p)^{k - 1}}p\).

04

Find the geometric distribution

(b)

The expected value of a geometric distribution is the reciprocal of the probability of success \(p\): \(\mu = \frac{1}{p}\).

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Most popular questions from this chapter

Question: Prove Theorem \(2\), the extended form of Bayes’ theorem. That is, suppose that \(E\) is an event from a sample space \(S\) and that \({F_1},{F_2},...,{F_n}\) are mutually exclusive events such that \(\bigcup\nolimits_{i = 1}^n {{F_i} = S} \). Assume that \(p\left( E \right) \ne 0\) and \(p\left( {{F_i}} \right) \ne 0\) for \(i = 1,2,...,n\). Show that

\(p\left( {{F_j}\left| E \right.} \right) = \frac{{p\left( {E\left| {{F_j}} \right.} \right)p\left( {{F_j}} \right)}}{{\sum\nolimits_{i = 1}^n {p\left( {E\left| {{F_i}} \right.} \right)p\left( {{F_i}} \right)} }}\)

(Hint: use the fact that \(E = \bigcup\nolimits_{i = 1}^n {\left( {E \cap {F_i}} \right)} \).)

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