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Chapter 6: Problem 17

Let \(X\) be the first time that a failure occurs in an infinite sequence of Bernoulli trials with probability \(p\) for success. Let \(p_{k}=P(X=k)\) for \(k=1,2, \ldots .\) Show that \(p_{k}=p^{k-1} q\) where \(q=1-p .\) Show that \(\sum_{k} p_{k}=1 .\) Show that \(E(X)=1 / q .\) What is the expected number of tosses of a coin required to obtain the first tail?

Short Answer

Expert verified
The expected number of tosses to get the first tail is \( \frac{1}{q} = \frac{1}{1-p} \).

Step by step solution

01

Understanding the Problem

We need to find the probability distribution, the total probability, and the expected value of the random variable representing the first failure in a sequence of Bernoulli trials.
02

Identifying Probability Distribution

The random variable \(X\) is the first failure in Bernoulli trials. The event \(X = k\) occurs if the first \(k-1\) trials are successes, each with probability \(p\), and the \(k\)-th trial is a failure with probability \(q = 1-p\). Thus: \[ p_k = P(X=k) = p^{k-1}q. \]
03

Confirming Total Probability is 1

The sum of all probabilities \(\sum_{k=1}^{\infty} p_k\) should be equal to 1. Since \(p_k = p^{k-1}q\), we have:\[ \sum_{k=1}^{\infty} p_{k} = \sum_{k=1}^{\infty} p^{k-1} q = q\sum_{k=0}^{\infty} p^k = q \frac{1}{1-p} = 1, \]which is correct since \(q = 1 - p\).
04

Determining the Expected Value

The expectation \(E(X)\) is the sum of each outcome multiplied by its probability:\[ E(X) = \sum_{k=1}^{\infty} k p_k = \sum_{k=1}^{\infty} k p^{k-1} q. \]Using the formula for the expected value of a geometric distribution, we derive:\[ E(X) = \frac{1}{q}, \]as expected of a geometric distribution.
05

Final Answer

Based on our calculations, the expected number of tosses required to obtain the first tail (failure in this case of a head being a success) is \( \frac{1}{q} \), which is \( \frac{1}{1-p} \) if heads are the success.

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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 one of the fundamental building blocks of probability and statistics. They are simple experiments where there are only two possible outcomes: success or failure. In this context, think of flipping a coin, where getting heads is considered a success and tails is a failure. Each flip is independent, meaning the outcome of one does not affect the outcome of another.
  • Success occurs with probability \( p \).
  • Failure occurs with probability \( q = 1 - p \).
When we talk about a sequence of Bernoulli trials, we are referring to multiple such coin flips happening in succession. The aim is often to find when the first failure (or success) occurs in this sequence.
Probability Distribution
In the context of Bernoulli trials, a probability distribution specifies how likely different outcomes are to happen. For our problem, we are interested in finding the probability that the first failure occurs at a particular point in the sequence.
This is modeled by a geometric distribution, where the Bernoulli trials stop when the first failure is observed. If we label the occurrence of failure as \( X \), then \( X = k \) means the first \( k-1 \) trials were successful, and the \( k \)-th is a failure. Hence, the probability \( p_k = p^{k-1}q \). This reflects the bernoulli successes followed by a single failure.
Expected Value
The expected value is a key concept in probability that provides the average outcome if an experiment is repeated many times. For the geometric distribution we're discussing, the expected value \(E(X)\) tells us the average number of trials needed before the first failure occurs.
For this geometric distribution:
  • The formula is \(E(X) = \frac{1}{q} \),
  • Where \( q = 1-p \) is the probability of failure.
In our coin toss example, if heads (a win) occurs with probability \( p \), then the expected number of tosses before the first tail (failure) is \( \frac{1}{1-p} \). This reveals how often you can expect to have to toss the coin to finally land tails.
Infinite Sequence
An infinite sequence implies that we're considering an endless series of Bernoulli trials. In theory, you could keep flipping a coin forever without encountering a failure. However, as the flips continue, the probability of encountering a failure all at once cannot surpass 1.
This means it is mathematically valid to calculate expectations and probabilities over an infinite series because, even though they never actually end:
  • The calculations converge to a finite number, which represents actual probabilities and expectations.
  • For example, the probability sum \( \sum_{k=1}^{\infty} p_k = 1 \) shows that eventually, a failure must occur within the infinite set of trials.
Total Probability
Total probability is about ensuring the sum of all possible outcomes equals 1, which confirms that all possible events have been considered. In the case of our geometric distribution, with outcomes \( p_k = p^{k-1}q \), the task is to show that this sum equals 1.
We confirm:
  • \( \sum_{k=1}^{\infty} p_k = q \sum_{k=0}^{\infty} p^k = q \frac{1}{1-p} = 1\).
  • This serves to reassure us that every possible scenario (each ordering of failures) is accounted for.
Thus, the total probability not only validates the distribution but also the methods used in defining the sequence of trials.

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Most popular questions from this chapter

A coin is tossed until the first time a head turns up. If this occurs on the \(n\) th toss and \(n\) is odd you win \(2^{n} / n,\) but if \(n\) is even then you lose \(2^{n} / n\). Then if your expected winnings exist they are given by the convergent series $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$ called the alternating harmonic series. It is tempting to say that this should be the expected value of the experiment. Show that if we were to do this, the expected value of an experiment would depend upon the order in which the outcomes are listed.

A long needle of length \(L\) much bigger than 1 is dropped on a grid with horizontal and vertical lines one unit apart. Show that the average number \(a\) of lines crossed is approximately $$a=\frac{4 L}{\pi}$$

In a class there are 20 students: 3 are \(5^{\prime} 6^{\prime \prime}, 5\) are \(5^{\prime} 8^{\prime \prime}, 4\) are \(5^{\prime} 10^{\prime \prime}, 4\) are 6 ', and 4 are \(6^{\prime} 2^{\prime \prime} .\) A student is chosen at random. What is the student's expected height?

A die is rolled twice. Let \(X\) denote the sum of the two numbers that turn up, and \(Y\) the difference of the numbers (specifically, the number on the first roll minus the number on the second). Show that \(E(X Y)=E(X) E(Y) .\) Are \(X\) and \(Y\) independent?

Let \(X\) be a random variable with range [-1,1] and density function \(f_{X}(x)=\) \(a x+b\) if \(|x|<1\) (a) Show that if \(\int_{-1}^{+1} f_{X}(x) d x=1,\) then \(b=1 / 2\). (b) Show that if \(f_{X}(x) \geq 0,\) then \(-1 / 2 \leq a \leq 1 / 2\). (c) Show that \(\mu=(2 / 3) a,\) and hence that \(-1 / 3 \leq \mu \leq 1 / 3\). (d) Show that \(\sigma^{2}(X)=(2 / 3) b-(4 / 9) a^{2}=1 / 3-(4 / 9) a^{2}\).
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