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A company that manufactures shoes has three factories. Factory 1 produces \(25 \%\) of the company's shoes, Factory 2 produces \(60 \%,\) and Factory 3 produces \(15 \%\) One percent of the shoes produced by Factory 1 are mislabeled, \(0.5 \%\) of those produced by Factory 2 are mislabeled, and \(2 \%\) of those produced by Factory 3 are mislabeled. If you purchase one pair of shoes manufactured by this company, what is the probability that the shoes are mislabeled?

Short Answer

Expert verified
The probability that a purchased pair of shoes is mislabeled is 0.0085.

Step by step solution

01

Identify the Probabilities

First, express all given percentages as probabilities: Factory 1 produces 0.25 of the company's shoes; Factory 2 produces 0.60; and Factory 3 produces 0.15. The corresponding mislabeling probabilities are 0.01, 0.005, and 0.02, respectively.
02

Calculate the Individual Probabilities

Next, calculate the probability of a pair of shoes being mislabeled from each factory. This is done by multiplying the production probability of each factory by their respective mislabeling probability. For Factory 1, this is \(0.25 * 0.01 = 0.0025\); For Factory 2, this is \(0.60 * 0.005 = 0.003\); And for Factory 3, this is \(0.15 * 0.02 = 0.003\).
03

Sum All Probabilities

To find the total probability of purchasing a mislabeled pair of shoes from the company, sum up all the calculated probabilities from step 2: \(0.0025 + 0.003 + 0.003 = 0.0085\).

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Key Concepts

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

Basic Probability
Understanding the concept of basic probability is the foundation for interpreting events and making predictions about outcomes in a random process. Probability, by definition, is the measure of how likely an event is to occur, with a value ranging between 0 and 1, where 0 signifies impossibility and 1 indicates certainty.

When analyzing the chance of a particular event happening, the process involves counting the number of favorable outcomes over the total number of possible outcomes. In the context of mislabeled shoes from a company with three factories, the probability of picking a mislabeled pair is understood as a favorable outcome (despite being unfavorable in practical terms). The basic probability formula used is:
\[ P(E) = \frac{\text{Number of ways event E can occur}}{\text{Total number of possible outcomes}} \]
However, in real-life applications, especially in manufacturing, you will often use given probabilities, like the likelihood of a factory producing a mislabeled pair of shoes, to calculate your final outcome.
Probability Calculations
Probability calculations involve using mathematical formulas to determine the likelihood of various events. The first step in solving a problem like the mislabeled shoes is often to break down a complex event into simpler, smaller events, each with its own probability, and then combine these to find the total probability.

In our example, individual probability calculations need to be performed for each factory. This requires multiplying the proportion of shoes each factory produces by the chance that an output from that factory is mislabeled.

Mathematical Expression of Individual Probabilities

For a pair of shoes from Factory 1, the calculation looks like: \[ P(\text{Mislabeled|Factory 1}) = P(\text{Factory 1}) \times P(\text{Mislabeled|Factory 1 produces}) \].
These individual probabilities are necessary for understanding the contribution of each part to the whole scenario, which is crucial for making accurate combined probability estimates.
Conditional Probability
Conditional probability is the likelihood of an event occurring given that another event has already occurred. In our shoes example, we are looking at the probability of a pair being mislabeled given that it was produced by a specific factory. This concept is paramount when dealing with dependent events, where the outcome or probability of one event affects another.

The formula for conditional probability is expressed as: \[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
where \( P(A|B) \) is the probability of event A given event B, and \( P(A \cap B) \) represents the probability of both events A and B happening. If events A and B are independent, then \( P(A \cap B) = P(A) \times P(B) \), just as we multiplied the factory's production probability by its mislabeling probability.
Percentage to Probability Conversion
In probability problems, percentages are often provided as data points, but for calculations, they need to be converted into decimal form. This is because probabilities are traditionally expressed as fractions or decimals, not percentages.

To convert a percentage to a probability, simply divide the percentage by 100. For example: \[ P(\text{Mislabeled|Factory 1}) = \frac{1}{100} = 0.01 \]
This is essential in the shoes manufacturing problem, where the probabilities of mislabeling are given in percentages and need to be converted as such for the purpose of calculation. Always ensuring that these conversions are correct is critical to finding the right overall probability.

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Most popular questions from this chapter

A woman and a man (unrelated) each has two children. At least one of the woman's children is a boy, and the man's older child is a boy. Is the probability that the woman has two boys greater than, equal to, or less than the probability that the man has two boys? a. Demonstrate the truth of your answer by using a simple sample to represent each family. b. Demonstrate the truth of your answer by taking two samples, one from men with two-children families and one from women with two-children families. c. Demonstrate the truth of your answer using computer simulation. Using the Bernoulli probability function with \(p=0.5 \text { (let } 0=\text { girl and } 1=\text { boy })\), generate 500 "families of two children" for the man and the woman. Determine which of the 500 satisfy the condition for each and determine the observed proportion with two boys. d. Demonstrate the truth of your answer by repeating the computer simulation several times. Repeat the simulation in part c several times. e. Do the preceding procedures scem to yicld the same results? Explain.

Consider the set of integers \(1,2,3,4,\) and 5 a. One integer is selected at random. What is the probability that it is odd? b. Two integers are selected at random (one at a time with replacement so that each of the five is available for a second selection). Find the probability that neither is odd; exactly one of them is odd; both are odd.

Suppose that \(\mathrm{A}\) and \(\mathrm{B}\) are events defined on a common sample space and that the following probabilities are known: \(P(\mathrm{A})=0.3, P(\mathrm{B})=0.4,\) and \(P(\mathrm{A} | \mathrm{B})=0.2\) Find \(P(\mathrm{A} \text { or } \mathrm{B})\)

\(P(\mathrm{R})=0.5, P(\mathrm{S})=0.3,\) and events \(\mathrm{R}\) and \(\mathrm{S}\) are independent. a. Find \(P(\mathrm{R} \text { and } \mathrm{S})\) b. \(\quad\) Find \(P(\mathrm{R} \text { or } \mathrm{S})\) c. Find \(P(\bar{S})\) d. \(\quad\) Find \(P(\mathbf{R} | \mathbf{S})\) e. Find \(P(\mathrm{S} | \mathrm{R})\) f. Are events \(R\) and \(S\) mutually exclusive? Explain.

A box contains 25 parts, of which 3 are defective and 22 are nondefective. If 2 parts are selected without replacement, find the following probabilities: a. \(\quad P(\text { both are defective })\) b. \(\quad P(\) exactly one is defective) c. \(\quad P(\text { neither is defective })\)

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