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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 that the next page you read contains (a) 0 typographical errors is approximately \(0.8187\), and (b) 2 or more typographical errors is approximately \(0.0176\). This is calculated using the Poisson distribution with an average rate of \(0.2\) errors per page.

Step by step solution

01

(Calculate the probability for scenario (a))

We'll use the Poisson probability mass function (PMF) to find the probability of 0 errors on a page. The PMF of the Poisson distribution is given by: \(P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!}\) where, \(X\) = number of events, \(k\) = desired number of events, in our case \(k=0\), \(\lambda\) = expected rate of events, in our case, \(\lambda = 0.2\), \(e\) = base of the natural logarithm, approximately equal to \(2.718\). Now, we substitute the values to find the probability for scenario (a): \(P(X = 0) = \frac{e^{-0.2} \cdot 0.2^0}{0!}\)
02

(Calculate the probability for scenario (b))

To calculate the probability of 2 or more errors, we'll first find the probabilities of 1 and 0 errors, and then subtract their sum from 1. This method is used because calculating the probabilities for 2, 3, 4, ... errors individually and then summing them up would be cumbersome. \(P(X \geq 2) = 1 - (P(X = 0) + P(X = 1))\) We already know \(P(X = 0)\), so now we just need to find \(P(X = 1)\). Using the Poisson PMF again, with \(k = 1\): \(P(X = 1) = \frac{e^{-0.2} \cdot 0.2^1}{1!}\) Finally, we plug the values in to find the probability of 2 or more errors: \(P(X \geq 2) = 1 - (P(X = 0) + P(X = 1))\)
03

(Compute the probabilities)

Now, we calculate the probabilities for scenarios (a) and (b). Scenario (a): \(P(X = 0) = \frac{e^{-0.2} \cdot 0.2^0}{0!} \approx 0.8187\) Scenario (b): First, calculate \(P(X = 1)\): \(P(X = 1) = \frac{e^{-0.2} \cdot 0.2^1}{1!} \approx 0.1637\) Then, find the probability of 2 or more errors: \(P(X \geq 2) = 1 - (P(X = 0) + P(X = 1)) \approx 1 - (0.8187 + 0.1637) \approx 0.0176\) Hence, the probability that the next page you read contains: (a) 0 typographical errors is approximately \(0.8187\). (b) 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.

Poisson Probability Mass Function
The Poisson probability mass function (PMF) is a formula used to determine the likelihood of a given number of events occurring within a fixed interval of time or space, where these events happen with a known average rate and independently of the time since the last event. For example, when assessing the probability of typing errors on a page, we can use the Poisson PMF if we know the average number of errors typically found on a page.

The function is expressed as: \[P(X = k) = \frac{e^{-\lambda} \cdot \lambda^k}{k!}\]

Here, \(e\) represents the base of the natural logarithm, \(\lambda\) is the expected number of events (which can be thought of as the event rate), and \(k\) is the number of occurrences for which we want to find the probability. The symbol \(!\) denotes factorial, which is the product of all positive integers up to that number. In our typographical error example, if the expected rate \(\lambda\) is 0.2 errors per page, using the PMF, we can calculate the probability of seeing exactly 0, 1, or more errors on a new page.
Expected Value
The expected value of a random variable is a fundamental concept in probability, representing the average outcome one would expect to see over a large number of trials or observations. In practical terms, it’s often regarded as the 'mean' or 'average' value expected. For a Poisson distribution, the expected value is equal to \(\lambda\), the rate parameter of the distribution.

In the context of typographical errors in a magazine, if one expects, on average, to find 0.2 errors per page, then the expected value \(\lambda\) is 0.2. This value not only sets the center or the 'peak' of the Poisson distribution but also reflects the dispersion since the variance of a Poisson distribution is also equal to \(\lambda\). Over many pages, we would expect an average of 0.2 errors per page, because that is our expected value.
Factorial
The factorial of a non-negative integer \(n\), denoted by \(n!\), is the product of all positive integers less than or equal to \(n\). It's a concept used in various areas of mathematics but is particularly important in probability and combinatorics. For instance, in the Poisson PMF, factorials appear in the denominator as a way to handle the various possible arrangements that could lead to the observed number of events.

Factorials grow at an extremely fast rate with larger values of \(n\). For example, \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\). In probability calculations, it is important when \(k\) equals zero, \(k!\) is defined to be 1, which plays a role when calculating the probability of zero events occurring, as in our case with \(P(X = 0)\) for typographical errors.
Poisson Distribution Applications
The Poisson distribution is incredibly versatile and can be applied to a wide range of practical situations, particularly those that involve counting occurrences of events. Common applications include calculating the number of calls received by a call center, the number of decay events from a radioactive source per unit time, and the number of vehicles passing through a traffic point.

The key characteristics for applying the Poisson distribution are situations where events occur randomly and independently within a given time or space, and there is a constant average rate of occurrence. In our exercise, calculating the number of typographical errors on a magazine page is just one of many instances where the Poisson distribution provides valuable insights into the likelihood of various outcomes based on an expected average rate (\(\lambda\)), thus enabling better decision-making and predictions in diverse fields such as telecommunications, healthcare, and quality control in production processes.

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Most popular questions from this chapter

A game popular in Nevada gambling casinos is Keno, which is played as follows: Twenty numbers are selected at random by the casino from the set of numbers 1 through 80. A player can select from 1 to 15 numbers; a win occurs if some fraction of the player's chosen subset matches with any of the 20 numbers drawn by the house. The payoff is a function of the number of elements in the player's selection and the number of matches. For instance, if the player selects only 1 number, then he or she wins if this number is among the set of 20 , and the payoff is \(\$ 2.2\) won for every dollar bet. (As the player's probability of winning in this case is \(\frac{1}{4}\), it is clear that the "fair" payoff should be \(\$ 3\) won for every \(\$ 1\) bet.) When the player selects 2 numbers, a payoff (of odds) of \(\$ 12\) won for every \(\$ 1\) bet is made when both numbers are among the 20 , (a) What would be the fair payoff in this case? Let \(P_{n, k}\) denote the probability that exactly \(k\) of the \(n\) numbers chosen by the player are among the 20 selected by the house. (b) Compute \(P_{n, k}\). (c) The most typical wager at Keno consists of selecting 10 numbers. For such a bet the casino pays off as shown in the following table. Compute the expected payoff:

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