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An accountant has observed that \(5 \%\) of all copies of a particular two-part form have an error in Part I, and \(2 \%\) have an error in Part II. If the errors occur independently, find the probability that a randomly selected form will be error-free.

Short Answer

Expert verified
The probability that the form is error-free is 0.931.

Step by step solution

01

Understand the Problem

We are given probabilities for errors in two parts of a form and need to find the probability that neither part has an error. The errors occur independently at the specified rates in each part.
02

Identify Given Probabilities

The probability that Part I has an error is given as \( P( ext{Error in Part I}) = 0.05 \), and the probability that Part II has an error is \( P( ext{Error in Part II}) = 0.02 \). These are independent events.
03

Calculate Probabilities of No Errors

To find the probability of no error in a part, take the complement of the probability of an error. \[ P( ext{No Error in Part I}) = 1 - 0.05 = 0.95 \] \[ P( ext{No Error in Part II}) = 1 - 0.02 = 0.98 \]
04

Calculate Probability of Both Parts Being Error-Free

Since the errors are independent, we multiply the probabilities of no errors in each part to find the total probability of the form being error-free.\[ P( ext{Error-Free}) = P( ext{No Error in Part I}) \times P( ext{No Error in Part II}) \]\[ P( ext{Error-Free}) = 0.95 \times 0.98 \]
05

Perform the Multiplication

Multiply the probabilities: \[ 0.95 \times 0.98 = 0.931 \]Thus, the probability that the form is error-free is 0.931.

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Key Concepts

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

Probability
Probability is a measure that quantifies the likelihood of an event occurring. It moves on a scale from 0 to 1, where 0 means the event will not happen, and 1 indicates certainty that it will occur. For example, if you flip a fair coin, the probability of it landing on heads is 0.5.
In problems involving independent events, like this one with form errors, probability helps us assess the chance of multiple conditions happening together. Here, the interest lies in ensuring a form is error-free in both parts, calculated by understanding the probability of errors and their complementary probabilities. Remember:
  • Independent events don't affect each other's outcomes.
  • The probability of all independent events occurring is found by multiplying their probabilities.
Understanding probability in this way provides clarity in determining outcomes when several events are involved.
Complement Rule
The complement rule is crucial for finding probabilities of events not happening. Simply stated, it says that the probability of an event not occurring is 1 minus the probability that it does occur. For instance, if the probability of having an error in Part I of a form is 0.05, then using the complement rule, the probability of not having an error is:\[P(\text{No Error in Part I}) = 1 - P(\text{Error in Part I}) = 1 - 0.05 = 0.95\]This concept is vital when calculating the probability of complementary and mutually exclusive events. Here, having no errors is the complement of having errors in each form part. Always perform a check by ensuring the total of an event and its complement equals 1:
  • Event Probability + Complement Probability = 1
This rule ensures correctness in our calculations. It lays the groundwork for finding the overall probability of independent events.
Multiplication Rule of Probability
The multiplication rule of probability is vital for calculating the probability of two independent events happening together. When two events are independent, the occurrence of one does not affect the occurrence of the other. Thus, the probability that both events occur is simply the product of their individual probabilities.
In our form error example, we calculate the chance of a form being error-free by multiplying the probabilities that each part of the form is error-free. This is shown as:\[P(\text{Error-Free}) = P(\text{No Error in Part I}) \times P(\text{No Error in Part II})\]Substituting the known probabilities:\[P(\text{Error-Free}) = 0.95 \times 0.98 = 0.931\]This calculation demonstrates how independent probabilities combine. By applying the multiplication rule, we find the joint probability of independent events—an essential step in multi-event probability scenarios.

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Most popular questions from this chapter

A random experiment gave rise to the two-way contingency table shown. Use it to compute the probabilities indicated. $$ \begin{array}{|l|l|l|} \hline & R & S \\ \hline A & 0.13 & 0.07 \\ \hline B & 0.61 & 0.19 \\ \hline \end{array} $$ a. \(P(A), P(R), P(A \cap R)\) b. Based on the answer to (a), determine whether or not the events \(A\) and \(R\) are independent. c. Based on the answer to (b), determine whether or not \(P(A \mid R)\) can be predicted without any computation. If so, make the prediction. In any case, compute \(P(A \mid R)\) using the Rule for Conditional Probability.

Make a statement in ordinary English that describes the complement of each event (do not simply insert the word "not"). a. In the roll of a die: "five or more." b. In a roll of a die: "an even number." c. In two tosses of a coin: "at least one heads." d. In the random selection of a college student: "Not a freshman."

The sample space that describes all three-child families according to the genders of the children with respect to birth order is $$ S=\\{b b b, b b g, b g b, b g g, g b b, g b g, g g b, g g g\\} $$ In the experiment of selecting a three-child family at random, compute each of the following probabilities, assuming all outcomes are equally likely. a. The probability that the family has at least two boys. b. The probability that the family has at least two boys, given that not all of the children are girls. c. The probability that at least one child is a boy. d. The probability that at least one child is a boy, given that the first born is a girl.

For the sample space \(\boldsymbol{S}-\\{a, b, c, d, e\\}\) identify the complement of each event given. a. \(A-\\{a, d, e\\}\) b. \(\boldsymbol{B}-\\{b, c, d, e\\}\) c. \(S\)

A man has two lights in his well house to keep the pipes from freezing in winter. He checks the lights daily. Each light has probability 0.002 of burning out before it is checked the next day (independently of the other light). a. If the lights are wired in parallel one will continue to shine even if the other burns out. In this situation, compute the probability that at least one light will continue to shine for the full 24 hours. Note the greatly increased reliability of the system of two bulbs over that of a single bulb. b. If the lights are wired in series neither one will continue to shine even if only one of them burns out. In this situation, compute the probability that at least one light will continue to shine for the full 24 hours. Note the slightly decreased reliability of the system of two bulbs over that of a single bulb.

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