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Chapter 4: Problem 48

It is known that diskettes produced by a certain company will be defective with probability \(.01,\) independently of each other. The company sells the diskettes in packages of size 10 and offers a money-back guarantee that at most 1 of the 10 diskettes in the package will be defective. The guarantee is that the customer can return the entire package of diskettes if he or she finds more than one defective diskette in it. If someone buys 3 packages, what is the probability that he or she will return exactly 1 of them?

Short Answer

Expert verified
The probability of returning exactly 1 of 3 packages can be calculated as follows: 1. Calculate the probability of having 0 and 1 defective disk: \\ P(X=0) = \(C(10, 0) \cdot (0.01)^0 \cdot (0.99)^{10}\) \\ P(X=1) = \(C(10, 1) \cdot (0.01)^1 \cdot (0.99)^{9}\) 2. Calculate the probability of having more than 1 defective disk in a package: \\ P(more than 1 defective disk) = 1 - P(0 defective disks) - P(1 defective disk) 3. Finally, calculate the probability of returning exactly 1 of 3 packages using the binomial probability formula: \\ P(return 1 package) = \(C(3, 1) \cdot p^1 \cdot (1-p)^{3-1}\) where p is the probability of having more than 1 defective disk in a package.

Step by step solution

01

Calculate the probability of a single defective disk

Since we know that the probability of a single disk being defective is 0.01, the probability of a disk not being defective is 1 - 0.01 = 0.99.
02

Calculate the probability of having exactly 1 defective disk in a package

To find this probability, we need to use the binomial probability formula: P(X = k) = \(C(n, k) \cdot p^k \cdot (1-p)^{n-k}\) where: - P(X = k) represents the probability of k successes (defective disks) in n trials (total number of disks) - C(n, k) is the total number of combinations of n items taken k at a time - p is the probability of success (probability of a disk being defective) - n is the total number of trials (total number of disks in a package) - k is the number of defective disks that we need Substitute the values n = 10, k = 1, and p = 0.01 into the formula: P(X = 1) = \(C(10, 1) \cdot (0.01)^1 \cdot (0.99)^{10-1}\) Calculate P(X = 1).
03

Calculate the probability of having more than 1 defective disk in one package

We can find the probability of having more than 1 defective disk in a package by subtracting the probability of having 0 or 1 defective disks from the total probability (1). P(more than 1 defective disk) = 1 - P(0 defective disks) - P(1 defective disk) First, calculate P(0 defective disks) using the same binomial probability formula with k = 0: P(X = 0) = \(C(10, 0) \cdot (0.01)^0 \cdot (0.99)^{10-0}\) Next, subtract P(0 defective disks) and P(1 defective disk) from 1 to find P(more than 1 defective disk).
04

Calculate the probability of returning exactly 1 of 3 packages

To find the probability of returning exactly 1 of 3 packages (meaning exactly one package has more than 1 defective disk), we can use the binomial probability formula with n = 3, k = 1, and p = P(more than 1 defective disk). P(return 1 package) = \(C(3, 1) \cdot p^1 \cdot (1-p)^{3-1}\) where p is the probability of having more than 1 defective disk in a package. Calculate P(return 1 package) to find the probability of returning exactly 1 of 3 packages.

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics concerned with analyzing random events and determining the likelihood of various outcomes. It's the foundation for predicting how probable it is that something will happen鈥攐r not happen. When we talk about an event's probability, we鈥檙e asking, 'What are the chances of this specific outcome occurring?'

For example, if a company knows that a diskette has a 0.01 probability of being defective, they鈥檙e saying that out of 100 diskettes, on average, one will be defective. This number is crucial in calculating risks and making decisions, like offering a money-back guarantee if too many defective products are found in a package.
Combinatorics
Combinatorics is a field of mathematics that deals with counting and arranging objects. It's integral to probability theory because it helps us figure out how many ways certain events can occur. The binomial probability formula, which is used to solve the exercise about defective diskettes, relies on combinatorics to calculate one of its parts: the total number of combinations, represented by the term C(n, k).

This 'combinations' part answers the question, 'In how many ways can you select k defective diskettes out of n total diskettes?' Understanding combinatorics is key to solving a wide range of probability problems.
Defective Product Probability
The probability of encountering a defective product, like a diskette, can be calculated using the binomial probability formula. The formula works under the assumption that each product has an independent chance of being defective. When a company offers a guarantee on their product, they need to understand this probability to avoid significant losses.

Using the formula, we can calculate the chance of different scenarios, such as no defective diskettes in a package, exactly one defective diskette, or more than one. The cost of refunds and returns hinges on these probabilities.
Probability of Independent Events
When events are independent, the outcome of one event doesn鈥檛 influence the outcome of another. In our example, each diskette's condition is independent of the others, meaning the probability of one diskette being defective does not affect the probability of another being defective.

This concept allows us to multiply probabilities of individual events to find the likelihood of several events occurring in sequence. It's also used in the binomial probability formula to calculate the chance of a specific number of independent successes (or failures) within a series of trials.

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Most popular questions from this chapter

A sample of 3 items is selected at random from a box containing 20 items of which 4 are defective. Find the expected number of defective items in the sample.

A total of \(2 n\) people, consisting of \(n\) married couples, are randomly seated (all possible orderings being equally likely) at a round table. Let \(C_{i}\) denote the event that the members of couple \(i\) are seated next to each other, \(i=1, \ldots, n\) (a) Find \(P\left(C_{i}\right)\) (b) For \(j \neq i,\) find \(P\left(C_{j} | C_{i}\right)\) (c) Approximate the probability, for \(n\) large, that there are no married couples who are seated next to each other.

Suppose that a die is rolled twice. What are the possible values that the following random variables can take on: (a) the maximum value to appear in the two rolls; (b) the minimum value to appear in the two rolls; (c) the sum of the two rolls; (d) the value of the first roll minus the value of the second roll?

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 \(p=\frac{1}{2}\)

The National Basketball Association (NBA) draft lottery involves the 11 teams that had the worst won-lost records during the year. A total of 66 balls are placed in an urn. Each of these balls is inscribed with the name of a team: Eleven have the name of the team with the worst record, 10 have the name of the team with the second-worst record, 9 have the name of the team with the thirdworst record, and so on (with 1 ball having the name of the team with the 11 th-worst record). A ball is then chosen at random, and the team whose name is on the ball is given the first pick in the draft of players about to enter the league. Another ball is then chosen, and if it "belongs" to a team different from the one that received the first draft pick, then the team to which it belongs receives the second draft pick. (If the ball belongs to the team receiving the first pick, then it is discarded and another one is chosen; this continues until the ball of another team is chosen.) Finally, another ball is chosen, and the team named on the ball (provided that it is different from the previous two teams) receives the third draft pick. The remaining draft picks 4 through 11 are then awarded to the 8 teams that did not "win the lottery," in inverse order of their won-lost records. For instance, if the team with the worst record did not receive any of the 3 lottery picks, then that team would receive the fourth draft pick. Let \(X\) denote the draft pick of the team with the worst record. Find the probability mass function of \(X\)
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