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Chapter 6: Problem 7

Consider a sequence of independent Bernoulli trials, each of which is a success with probability \(p\). Let \(X_{1}\) be the number of failures preceding the first success, and let \(X_{2}\) be the number of failures between the first two successes. Find the joint mass function of \(X_{1}\) and \(X_{2}\)

Short Answer

Expert verified
The joint mass function of \(X_1\) and \(X_2\) is \(P(X_1 = x_1, X_2 = x_2) = p^2 (1-p)^{x_1 + x_2}\).

Step by step solution

01

Recall the probability function for Bernoulli trials

Recall that the probability mass function of a geometric distribution is given by: \(P(X = x) = p(1-p)^x\) Where \(p\) is the probability of success and \(x\) is the number of failures.
02

Define the variables of our problem

Let's denote the following variables for ease of representation: \(X_1\) = number of failures before the first success, \(X_2\) = number of failures between the first two successes, and \(p\) = probability of success.
03

Compute the joint mass function

The probability of having \(x_1\) failures before the first success and \(x_2\) failures between the first two successes can be computed as: \(P(X_1 = x_1, X_2 = x_2)\) Since the trials are independent, the probability of each event happening is the product of the probabilities for each event: \(P(X_1 = x_1, X_2 = x_2) = P(X_1 = x_1) * P(X_2 = x_2)\) Now, we can substitute the probability mass function for geometric distribution for both \(P(X_1 = x_1)\) and \(P(X_2 = x_2)\): \(P(X_1 = x_1, X_2 = x_2) = p(1-p)^{x_1} * p(1-p)^{x_2}\) Then, simplify by combining the terms: \(P(X_1 = x_1, X_2 = x_2) = p^2 (1-p)^{x_1 + x_2}\) Thus, the joint mass function of \(X_1\) and \(X_2\) is: \(P(X_1 = x_1, X_2 = x_2) = p^2 (1-p)^{x_1 + x_2}\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Bernoulli Trials
Bernoulli trials are a fundamental concept in probability theory, named after the Swiss mathematician Jacob Bernoulli. They represent experiments or processes that produce just two possible outcomes: a success (with probability p) or a failure (with probability 1-p). A common example of a Bernoulli trial is a coin toss, where the probability of landing heads (success) or tails (failure) are typically both 0.5 if the coin is fair.

One of the unique characteristics of Bernoulli trials is that each trial is independent of the others. This means that the outcome of any given trial does not affect the outcomes of the subsequent trials. As part of our educational initiative, understanding Bernoulli trials sets a solid foundation for more complex probability distributions such as the geometric distribution, which derives directly from these trials.
Geometric Distribution
The geometric distribution is an important probability distribution that describes the number of Bernoulli trials needed before a success occurs. It is directly connected to Bernoulli trials in that it models the 'waiting time' until a success. A discrete random variable X that is geometrically distributed reflects the failures we observe before we hit our first success.

For example, in a series of coin tosses, if we are looking for the first heads to occur, the geometric distribution tells us the probability of seeing a certain number of tails before we get that heads. It's worth mentioning that the term 'memoryless' is often associated with this distribution, meaning that past outcomes don't affect the probability of the outcome of future trials.
Probability Mass Function
A probability mass function (PMF) is a function that gives us the probability that a discrete random variable is exactly equal to some value. For Bernoulli trials, the geometric distribution has a PMF described by the formula
\(P(X = x) = p(1-p)^x\)

where x is the number of failures and p is the probability of a success on each trial. The PMF is crucial for determining the likelihood of different outcomes in a discrete space. It's the cornerstone of solving problems involving discrete random variables, as it maps the probabilities for all possible outcomes.
Independent Random Variables
Independent random variables are a central concept in probability and statistics. When two variables are independent, the occurrence of one event does not influence the probability of the other event. In the case of our exercise scenario, having X1 represent the number of failures before the first success and X2 representing the number of failures between the first two successes is illustrative of this independence.

Independence allows us to multiply their respective probabilities to obtain the joint mass function since the occurrence of failures for X1 does not at all affect the trials for X2. This property simplifies the computation of the joint probability of multiple random variables and is vital in understanding how complex systems behave when individual components operate independently.

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Most popular questions from this chapter

A television store owner figures that 45 percent of the customers entering his store will purchase an ordinary television set, 15 percent will purchase a color television set, and 40 percent will just be browsing. If 5 customers enter his store on a given day, what is the probability that he will sell exactly 2 ordinary sets and 1 color set on that day?

Choose a number \(X\) at random from the set of numbers \(\\{1,2,3,4,5\\}\). Now choose a number at random from the subset no larger than \(X\), that is, from \(\\{1, \ldots, X]\). Call this second number \(Y\). (a) Find the joint mass function of \(X\) and \(Y\). (b) Find the conditional mass function of \(X\) given that \(Y=i\). Do it for \(i=1,2,3,4,5 .\) (c) Are \(X\) and \(Y\) independent? Why?

Suggest a procedure for using Buffon's needle problem to estimate \(\pi\). Surprisingly enough, this was once a common method of evaluating \(\pi\).

The joint density function of \(X\) and \(Y\) is $$ f(x, y)= \begin{cases}x+y & 0

Consider a directory of classified advertisements that consists of \(m\) pages, where \(m\) is very large. Suppose that the number of advertisements per page varies and that your only method of finding out how many advertisements there are on a specified page is to count them. In addition, suppose that there are too many pages for it to be feasible to make a complete count of the total number of advertisements and that your objective is to choose a directory advertisement in such a way that each of them has an equal chance of being selected. (a) If you randomly choose a page and then randomly choose an advertisement from that page, would that satisfy your objective? Why or why not? Let \(n(i)\) denote the number of advertisements on page \(i, i=1, \ldots, m\), and suppose that whereas these quantities are unknown, we can assume that they are all less than or equal to some specified value \(n\). Consider the following algorithm for choosing an advertisement. Step 1. Choose a page at random. Suppose it is page \(X\). Determine \(n(X)\) by counting the number of advertisements on page \(X\). Step 2. "Accept" page \(X\) with probability \(n(X) / n\). If page \(X\) is accepted, go to step 3. Otherwise, return to step 1 . Step 3. Randomly choose one of the advertisements on page \(X\).
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