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91Ó°ÊÓ

A certified public accountant (CPA) has found that nine of ten company audits contain substantial errors. If the CPA audits a series of company accounts, what is the probability that the first account containing substantial errors a. is the third one to be audited? b. will occur on or after the third audited account?

Short Answer

Expert verified
a) 0.009; b) 0.01

Step by step solution

01

Define the Probability of Errors

The probability that a single audit contains a substantial error is 9 out of 10, or \( P(\text{error}) = 0.9 \). Conversely, the probability of an account being error-free is \( P(\text{no error}) = 0.1 \).
02

Probability of First Error on the Third Audit

To find the probability that the first error occurs on the third audit, we need two accounts to be error-free, followed by one account with an error. This can be represented with the sequence "no error, no error, error." The probability is calculated as \( P(\text{first error on third}) = (0.1)^2 \times 0.9 \).
03

Compute the Probability for Part (a)

Calculating the probability: \( P(\text{first error on third}) = 0.1 \times 0.1 \times 0.9 = 0.009 \).
04

Probability of First Error On or After Third Audit

To find the probability that the first error occurs on or after the third audit, we recognize that this is the complement of having the first error in either the first or the second audit. Calculate \( P(\text{no error in first two audits}) = (0.1)^2 = 0.01 \). Thus, \( P(\text{error on or after third audit}) = 1 - P(\text{error in first two}) \).
05

Compute the Probability for Part (b)

Calculate \( P(\text{error on or after third audit}) = 1 - (0.9 + 0.09) = 1 - 0.99 = 0.01 \).

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Key Concepts

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

Binomial Distribution
The **binomial distribution** is a probability distribution that sums up the results of a sequence of independent trials, where each trial can have just two possible outcomes: success or failure. These trials are identical and follow the same probability on each run. For example, when performing multiple audits, as in our case, each audit can result in "error" (success for our probability scenario) or "no error" (failure).
In our exercise, though, we don't directly use the full binomial formula, it's relevant when errors follow a specific pattern focus like seeing exactly a certain number of audits with errors within a fixed total number of audits.
Geometric Distribution
The **geometric distribution** tells us the probability that the first occurrence of a success will happen on a specific trial. This is particularly relevant for part (a) of our problem, where we want to know the likelihood of the first error appearing on the third audit.
The geometric distribution is applicable when we have a constant probability of success, here given as 0.9 for "error", across trials. The probability that the first "success" (error) appears on the third attempt can be calculated by multiplying the probabilities of failure (no error, 0.1) occurring in the first two audits, followed by a "success" (error) in the third audit:
  • Two no-errors: (0.1) and (0.1)
  • Followed by an error: (0.9)
This calculation gives us the likelihood of the discussed scenario.
Complement Rule
The **complement rule** is a simple yet powerful concept in probability that helps in finding the probability of an event occurring by subtracting the probability of the complementary event from 1.
In step 4 of our solution, we apply this rule to determine the probability that the first error occurs on or after the third audit.
Instead of calculating this directly, which can be complex, we find the probability of errors occurring in the first two audits and then subtract this from 1. This way, it simplifies finding the likelihood of the error not happening within the first two attempts, as "on or after third" includes everything else not in those initial trials.
Error Probability
**Error probability** in this context refers to the likelihood that an audit will result in finding substantial errors. The exercise specifies that the probability of an audit containing errors is exceptionally high, at 0.9 for every audit conducted.
Error probability helps to set the foundation for the geometric and complement rule applications. Understanding error probability is crucial as it impacts every calculation, especially in assessing how often errors are detected within the audits. Calculating the opposite of this figure, as shown with the probability of no-error being 0.1, is equally essential.
  • This smaller probability aids in determining unbroken sequences of error-free audits.
  • Allows easier calculation of more complex scenarios using complement rule.
By grounding probability calculations in this base understanding, various probable outcomes are more readily assessed.

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Most popular questions from this chapter

Suppose that \(Y\) is a binomial random variable based on \(n\) trials with success probability \(p\) and consider \(Y^{*}=n-Y\) a. Argue that for \(y^{*}=0,1, \ldots, n\) $$ P\left(Y^{*}=y^{*}\right)=P\left(n-Y=y^{*}\right)=P\left(Y=n-y^{*}\right) $$ b. Use the result from part (a) to show that $$ P\left(Y^{*}=y^{*}\right)=\left(\begin{array}{c} n \\ n-y^{*} \end{array}\right) p^{n-y^{*}} q^{y^{*}}=\left(\begin{array}{c} n \\ y^{*} \end{array}\right) q^{y^{*}} p^{n-y^{*}} $$ c. The result in part (b) implies that \(Y^{*}\) has a binomial distribution based on \(n\) trials and "success" probability \(p^{*}=q=1-p .\) Why is this result "obvious"?

A lot of \(N=100\) industrial products contains 40 defectives. Let \(Y\) be the number of defectives in a random sample of size 20. Find \(p(10)\) by using the hypergeometric probability distribution and the binomial probability distribution. Is \(N\) large enough that the value for \(p(10)\) obtained from the binomial distribution is a good approximation to that obtained using the hypergeometric distribution?

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