/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 20 An advertiser drops 10,000 leafl... [FREE SOLUTION] | 91Ó°ÊÓ

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An advertiser drops 10,000 leaflets on a city which has 2000 blocks. Assume that each leaflet has an equal chance of landing on each block. What is the probability that a particular block will receive no leaflets?

Short Answer

Expert verified
The probability that a particular block will receive no leaflets is approximately 0.0067.

Step by step solution

01

Define Probability

We first need to determine the probability of a single leaflet landing on a particular block. Since there are 2000 blocks and each leaflet has an equal chance, the probability of a leaflet landing on any given block is \( \frac{1}{2000} \).
02

Calculate Probability of No Leaflets

Given that each leaflet lands independently, the probability that a particular block receives no leaflets from one leaflet is \( 1 - \frac{1}{2000} \). Since there are 10,000 leaflets, and each one can potentially land on this block independently, we raise this probability to the 10,000th power: \( \left( 1 - \frac{1}{2000} \right)^{10000} \).
03

Approximate Using Exponential Limit

For large \( n \), \( \left(1 - \frac{1}{n}\right)^n \approx e^{-1} \), so \( \left(1 - \frac{1}{2000}\right)^{10000} \approx e^{-5} \). This simplification uses the fact that \( 10000 \times \frac{1}{2000} = 5 \).
04

Use Exponential Approximation

Using the approximation for \( e^{-5} \), which is about 0.0067, we find that the probability that a particular block receives no leaflets is approximately 0.0067.

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Key Concepts

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

Exponential Limit
In probability theory, exponential limits are a fascinating concept that can help make calculations easier. For instance, when finding the probability of an event under certain conditions, we sometimes encounter expressions like \( \left(1 - \frac{1}{n}\right)^n \) when \( n \) becomes large. With this, we can use an approximation where \( \left(1 - \frac{1}{n}\right)^n \approx e^{-1} \). This means that as \( n \) approaches infinity, the expression approaches the value of approximately 0.3679. Using this exponential limit helps simplify problems involving large numbers and allows us to work with more manageable figures.
This approximation is particularly useful when you need to find probabilities involving large datasets or numerous events, such as the 10,000 leaflets in our problem.
Independent Events
Understanding independent events is key to probability calculations. In simple terms, two events are independent if the outcome of one does not affect the other. In our example, the landing of one leaflet on a city block does not influence the landing of another.
When events are independent, we can calculate the total probability by multiplying the probabilities of each event. For instance, the probability of one leaflet landing on a block is \( \frac{1}{2000} \). Thus, the probability of none of the 10,000 independent leaflets landing on that block becomes \( \left(1 - \frac{1}{2000}\right)^{10000} \). This simplification is crucial for tackling complex probability problems without letting them become unmanageable.
Probability Calculation
Probability calculation is the backbone of understanding how likely events are to occur. Using basic principles, you can determine outcomes for any scenario.
To calculate the probability of no leaflets landing on a block, you start by identifying the basic probability of one event happening: a single leaflet landing. This is \( \frac{1}{2000} \). The complement, \( 1 - \frac{1}{2000} \), is the probability that it does not land on the block. With 10,000 leaflets, this probability must be applied to every single leaflet. Hence, we use exponential forms: \( \left(1 - \frac{1}{2000}\right)^{10000} \). This results in understanding the probability of the block receiving no leaflets at all, combining the independent probabilities into a single figure.
Approximation Methods
Approximation methods are invaluable in probability theory, especially for simplifying calculations when dealing with large quantities. As seen in our exercise, using approximations like the exponential limit allows us to avoid tedious calculations.
In practice, you assess complex expressions and replace them with simpler, near-equivalent values. For example, instead of laboriously computing \( \left(1 - \frac{1}{2000}\right)^{10000} \), we can use the approximation \( e^{-5} \), which is 0.0067. This method highlights how probability theory leverages mathematical properties for more efficient problem-solving. Approximations don't just save time—they make complicated probability problems approachable and provide a clearer understanding of underlying patterns.

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Most popular questions from this chapter

A die is rolled until the first time \(T\) that a six turns up. (a) What is the probability distribution for \(T ?\) (b) Find \(P(T>3)\). (c) Find \(P(T>6 \mid T>3)\).

There are an unknown number of moose on Isle Royale (a National Park in Lake Superior). To estimate the number of moose, 50 moose are captured and tagged. Six months later 200 moose are captured and it is found that 8 of these were tagged. Estimate the number of moose on Isle Royale from these data, and then verify your guess by computer program (see Exercise 36 ).

(a) Compute the leading digits of the first 100 powers of 2 , and see how well these data fit the Benford distribution. (b) Multiply each number in the data set of part (a) by \(3,\) and compare the distribution of the leading digits with the Benford distribution.

You are presented with four different dice. The first one has two sides marked 0 and four sides marked \(4 .\) The second one has a 3 on every side. The third one has a 2 on four sides and a 6 on two sides, and the fourth one has a 1 on three sides and a 5 on three sides. You allow your friend to pick any of the four dice he wishes. Then you pick one of the remaining three and you each roll your die. The person with the largest number showing wins a dollar. Show that you can choose your die so that you have probability \(2 / 3\) of winning no matter which die your friend picks. (See Tenney and Foster. \(\left.^{8}\right)\)

Let \(X\) be a random variable having a normal density and consider the random variable \(Y=e^{X}\). Then \(Y\) has a log normal density. Find this density of \(Y\).

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