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The expected number of typographical errors on a page of a certain magazine is .2. What is the probability that the next page you read contains (a) 0 and (b) 2 or more typographical errors? Explain your reasoning!

Short Answer

Expert verified
The probability for the next page you read containing (a) 0 typographical errors is approximately 0.8187, and (b) 2 or more typographical errors is approximately 0.0176. We used the Poisson distribution to calculate these probabilities, as it models the discrete probability of a given number of rare events (typographical errors) occurring in a fixed interval (one page).

Step by step solution

01

Understand the Poisson distribution and its parameters

The Poisson distribution is a discrete probability distribution that expresses the probability of a given number of events (in this case, typographical errors) occurring in a fixed interval. The probability mass function of the Poisson distribution is given by: \[P(k;\lambda) = e^{-\lambda}\frac{\lambda^k}{k!}\] Where: - \(k\) is the number of events (typographical errors), - \(\lambda\) is the average number of events in the interval (the expected number of typographical errors per page, which is 0.2 in this case), - \(e\) is the base of the natural logarithm (approximately equal to 2.71828).
02

Calculate the probability of 0 typographical errors on the next page

To find the probability that the next page we read contains 0 typographical errors, we will use the Poisson distribution formula with \(k = 0\): \(P(0; 0.2) = e^{-0.2}\frac{0.2^0}{0!}\) Since \(0! = 1\) and \(0.2^0 = 1\), then: \(P(0; 0.2) = e^{-0.2}\) Calculate the value of the probability: \(P(0; 0.2) \approx 0.8187\) Therefore, the probability that the next page we read contains 0 typographical errors is approximately 0.8187.
03

Calculate the probability of 2 or more typographical errors on the next page

To find the probability that the next page we read contains 2 or more typographical errors, we will use the Poisson distribution formula and the complement rule. First, we need to find the probability of 1 or fewer errors (i.e., 0 or 1 errors), then subtract that probability from 1. Calculate the probability of 1 error: \(P(1; 0.2) = e^{-0.2}\frac{0.2^1}{1!}\) \(P(1; 0.2) = 0.2e^{-0.2}\) \(P(1; 0.2) \approx 0.1637\) Now, use the complement rule to get the probability of 2 or more errors: \(P(2 \text{ or more}; 0.2) = 1 - (P(0; 0.2) + P(1; 0.2))\) \(P(2 \text{ or more}; 0.2) \approx 1 - (0.8187 + 0.1637)\) \(P(2 \text{ or more}; 0.2) \approx 0.0176\) Therefore, the probability that the next page we read contains 2 or more typographical errors is approximately 0.0176.

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Key Concepts

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

Probability Mass Function
The probability mass function (PMF) is a fundamental concept in understanding discrete probability distributions, like the Poisson distribution from our exercise. Essentially, the PMF describes how the probability is distributed over the possible numbers of events. For a random variable X that takes on a set of possible values, the PMF, represented as P(X=k), gives the probability that X equals a specific value k.

For instance, when we talk about typographical errors on a page, we can only have whole numbers of errors - you can't have half an error! That's why it's a discrete distribution, with each possible number of errors having a certain probability assigned by the PMF. In the case of Poisson distribution, the PMF is particularly useful for events that occur independently and at a constant rate within a given continuous interval, like typographical errors on pages.
Discrete Probability Distribution
In contrast to continuous distributions, discrete probability distributions apply to scenarios where outcomes can only take certain specific values. This is often the case when you're counting occurrences, just like typographical errors in our textbook exercise.

A discrete distribution has a probability mass function that can assign a probability to each potential outcome. Since these probabilities must sum up to 1, working out the likelihood of any and all possible scenarios is straightforward. The Poisson distribution is one such discrete probability distribution that predicts the probability of a given number of events occurring over a specified interval, assuming these events happen at a constant rate and are independent of each other's timings.
Factorial in Probability
When dealing with probability, particularly in the computation of discrete distributions such as the Poisson distribution, the concept of factorial is vital. Symbolized by an exclamation point (!), the factorial of a non-negative integer n, denoted by n!, is the product of all positive integers less than or equal to n.

For example, the factorial of 3 is calculated as: 3! = 3 × 2 × 1 = 6.In the context of our exercise, the factorial appears in the denominator of the Poisson probability mass function. It's part of what makes the distribution discrete, as it only accepts integer values for the number of events. As such, understanding factorials is crucial for calculating probabilities using the Poisson formula.
Complement Rule in Probability
The complement rule is a simple but powerful concept in probability theory. It states that the probability of an event not occurring is equal to one minus the probability of the event occurring. In formal terms, if the probability of event A is P(A), then the probability of not A, written as P(A'), is 1 - P(A).

Applying the complement rule helps to simplify complex calculations, especially in situations where it's easier to calculate the chance of something not happening than the opposite scenario. In our textbook example, we used the complement rule to find the probability of there being two or more errors by subtracting the probabilities of there being zero or one error from one. This rule is a cornerstone of probability theory and shows up in various forms across different problems and distributions.

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Most popular questions from this chapter

\(A\) and \(B\) play the following game: \(A\) writes down either number 1 or number \(2,\) and \(B\) must guess which one. If the number that \(A\) has written down is \(i\) and \(B\) has guessed correctly, \(B\) receives \(i\) units from \(A\). If \(B\) makes a wrong guess, \(B\) pays \(\frac{3}{4}\) unit to \(A .\) If \(B\) randomizes his decision by guessing 1 with probability \(p\) and 2 with probability \(1-p,\) determine his expected gain if (a) \(A\) has written down number 1 and (b) \(A\) has written down number 2 What value of \(p\) maximizes the minimum possible value of \(B\) 's expected gain, and what is this maximin value? (Note that \(B\) 's expected gain depends not only on \(p,\) but also on what \(A\) does.) Consider now player \(A\). Suppose that she also randomizes her decision, writing down number 1 with probability q. What is \(A\) 's expected loss if (c) \(B\) chooses number 1 and (d) \(B\) chooses number \(2 ?\) What value of \(q\) minimizes \(A\) 's maximum expected loss? Show that the minimum of \(A\) 's maximum expected loss is equal to the maximum of \(B\) 's minimum expected gain. This result, known as the minimax theorem, was first established in generality by the mathematician John von Neumann and is the fundamental result in the mathematical discipline known as the theory of games. The common value is called the value of the game to player \(B\).

Suppose that 10 balls are put into 5 boxes, with each ball independently being put in box \(i\) with probability \(p_{i}, \sum_{i=1}^{5} p_{i}=1\) (a) Find the expected number of boxes that do not have any balls. (b) Find the expected number of boxes that have exactly 1 ball.

Consider \(n\) independent trials, each of which results in one of the outcomes \(1, \ldots, k\) with respective probabilities \(p_{1}, \ldots, p_{k}, \quad \sum_{i=1}^{k} p_{i}=1 .\) Show that if all the \(p_{i}\) are small, then the probability that no trial outcome occurs more than once is approximately equal to \(\exp (-n(n-1)\) \(\left.\sum_{i} p_{i}^{2} / 2\right)\).

A gambling book recommends the following "winning strategy" for the game of roulette: Bet \(\$ 1\) on red. If red appears (which has probability \(\frac{18}{38}\) ), then take the \(\$ 1\) profit and quit. If red does not appear and you lose this bet (which has probability \(\frac{20}{38}\) of occurring), make additional \(\$ 1\) bets on red on each of the next two spins of the roulette wheel and then quit. Let \(X\) denote your winnings when you quit. (a) Find \(P\\{X > 0\\}\) (b) Are you convinced that the strategy is indeed a "winning" strategy? Explain your answer! (c) Find \(E[X]\)

Suppose that two teams play a series of games that ends when one of them has won \(i\) games. Suppose that each game played is, independently, won by team \(A\) with probability \(p .\) Find the expected number of games that are played when (a) \(i=2\) and (b) \(i=3 .\) Also, show in both cases that this number is maximized when \(p=\frac{1}{2}\).

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