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Chapter 3: Problem 4

An entrepreneur owns six corporations, each with more than \(\$ 10\) million in assets. The entrepreneur consults the U.S. Internal Revenue Data Book and discovers that the IRS audits \(15.3 \%\) of businesses of that size. What is the probability that two or more of these businesses will be audited?

Short Answer

Expert verified
To find the probability that two or more corporations will be audited, we firstly need to calculate the binomial probabilities of no audits and exactly one audit. We then subtract these probabilities from 1 to get the probability of two or more audits.

Step by step solution

01

Calculate probability of no audit

For this step, we use the formula for binomial probability which is given by \(P(X=k)= C(n, k) * (p^{k}) * (1-p)^{n-k}\), where \(n\) is the total number of trials (corporations), \(k\) is the desired number of successes (number of audits), \(p\) is the probability of success, and \(C(n, k)\) is the number of combinations of \(n\) items taken \(k\) at a time. Here, \(n = 6\), the number of corporations, and \(p = 0.153\), the probability a corporation gets audited. Let's find out the probability of no corporations getting audited, \(P(X=0)\). We plug in these values to get: \(P(X=0) = C(6,0) * (0.153^0) * (1-0.153)^{6}\).
02

Calculate probability of exactly one audit

Now let's find the probability of exactly one corporation being audited, i.e., \(P(X=1)\). We substitute \(n\), \(p\), and \(k=1\) in the binomial probability formula to get: \(P(X=1) = C(6,1) * (0.153^1) * (1-0.153)^{6-1}\).
03

Calculate probability of two or more audits

The probability of two or more corporations being audited is given by \(1 - P(X=0) - P(X=1)\), i.e., one minus the sum of the probabilities of no audits and exactly one audit.

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Key Concepts

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

Binomial Probability
To understand binomial probability, imagine flipping a coin. Each flip has two possible outcomes: heads or tails. Similarly, in situations where you have two possible outcomes for each trial, such as being audited or not, you can use the concept of binomial probability.

Binomial probability is calculated using the formula:

\[P(X=k) = C(n, k) \times (p^k) \times (1-p)^{n-k}\]
Here, \(n\) represents the number of independent trials, \(k\) is the number of successful outcomes we're interested in, \(p\) is the probability of success on a single trial, and \(C(n, k)\) is the number of combinations. In the exercise, the entrepreneur is assessing the likelihood of two or more audits, which requires us to look at multiple possibilities and add them up鈥攐r in this case, subtract them from 1 to get the complement.
Probability of Success
In any probability scenario, defining the 'probability of success' is key. It is the chance that a specific outcome we desire will occur. It can range from 0 (impossible event) to 1 (certain event). Understanding this concept allows you to set realistic expectations for the outcome.

For the entrepreneur, the 'success' in this context is an undesirable one鈥攂eing audited by the IRS. With the given audit probability of 15.3%, or \(0.153\), for a business with more than $10 million in assets, calculating the probabilities for different numbers of audits becomes straightforward using the binomial formula.
Combinations in Statistics
When it comes to statistics, the concept of combinations is crucial. Combinations involve selecting items from a group where the order of selection does not matter. This is represented by the notation \(C(n, k)\), or sometimes \({n \choose k}\), and is used in our binomial probability formula.

The mathematical formula for the number of combinations is:
\[C(n, k) = \frac{n!}{k!(n-k)!}\]
where \(n!\) denotes the factorial of \(n\), a product of all positive integers up to \(n\). For instance, when calculating the probability of no audits \(P(X=0)\), \(C(6, 0)\) simply equals 1, since there is only one way to choose zero items from six. For one audit \(P(X=1)\), \(C(6, 1)\) equals 6, representing the six possible scenarios where any one of the six corporations could be the one audited.

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Most popular questions from this chapter

A city has 74,806 registered automobiles. Each is required to display a bumper decal showing that the owner paid an annual wheel tax of \(\$ 50\). By law, new decals need to be purchased during the month of the owner's birthday. How much wheel tax revenue can the city expect to receive in November?

A gambler plays \(n\) hands of poker. If he wins the \(k\) th hand, he collects \(k\) dollars; if he loses the \(k\) th hand, he collects nothing. Let \(T\) denote his total winnings in \(n\) hands. Assuming that his chances of winning each hand are constant and independent of his success or failure at any other hand, find \(E(T)\) and \(\operatorname{Var}(T)\).

Suppose that \(r\) of \(N\) chips are red. Divide the chips into three groups of sizes \(n_{1}, n_{2}\), and \(n_{3}\), where \(n_{1}+\) \(n_{2}+n_{3}=N\). Generalize the hypergeometric distribution to find the probability that the first group contains \(r_{1}\) red chips, the second group \(r_{2}\) red chips, and the third group \(r_{3}\) red chips, where \(r_{1}+r_{2}+r_{3}=r\).

. A tool and die company makes castings for steel stress-monitoring gauges. Their annual profit, \(Q\), in hundreds of thousands of dollars, can be expressed as a function of product demand, \(y\) : $$ Q(y)=2\left(1-e^{-2 y}\right) $$ Suppose that the demand (in thousands) for their castings follows an exponential pdf, \(f_{Y}(y)=6 e^{-6 y}, y>0\). Find the company's expected profit.

For the St. Petersburg problem (Example 3.5.5), find the expected payoff if (a) the amounts won are \(c^{k}\) instead of \(2^{k}\), where \(0<\) \(c<2\). (b) the amounts won are \(\log 2^{k}\). [This was a modification suggested by D. Bernoulli (a nephew of James Bernoulli) to take into account the decreasing marginal utility of money \(-\) the more you have, the less useful a bit more is.]
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