Question

Tax returns audited by the IRS. According to the Internal Revenue Service (IRS), the chances of your tax return being audited are about 1 in 100 if your income is less than \(1 million and 9 in 100 if your income is \)1 million or more (IRS Enforcement and Services Statistics).

1. What is the probability that a taxpayer with income less than \(1 million will be audited by the IRS with income \)1 million or more?
2. If five taxpayers with incomes under \(1 million are randomly selected, what is the probability that exactly one will be audited? That more than one will be audited?
3. Repeat part b, assuming that five taxpayers with incomes of \)1 million or more are randomly selected.
4. If two taxpayers with incomes under \(1 million are randomly selected, and two with incomes more than \)1 million are randomly selected, what is the probability that none of these taxpayers will be audited by the IRS?
5. What assumptions did you have to make in order to answer these questions using the methodology presented in this section?

Short answer

1. The probability that taxpayers with income less than $1 million are 0.01, and the probability that taxpayers with income of $ 1 or more is 0.09.

2. The probability that exactly one will be audited by the IRS is 0.001.

3. The probability that exactly one will be audited by the IRS is 0.3068, and the probability more than one will be audited by the IRS is 0.0674.

4. The probability that none of these taxpayers will be audited by the IRS is 0.8116.

5. The required assumptions are:

The events are independent.

The probability of success is equal in all the trials.

Step-by-step solution

Given information

According to the Internal Revenue Service (IRS), if income is less than $1 million, the probability of tax returns being audited is about 1 in 100. If income is $1 million or more, the probability of tax return is 9 in 100.

(a) Calculating the probability that the IRS will audit a taxpayer with income less than $1 million

Let x denotes the taxpayers with income less than $1 million.

Let y denote the taxpayer with income equal to $1 million or more.

From the given information, the probability that a taxpayer with an income less than $1 million is given by,

$$\begin{gathered} \mathrm{p}\left(\mathrm{x}<1\right)=\frac{1}{100} \\ =0.01 \end{gathered}$$

Also, the probability that taxpayer with income equal to $1 million or more is given by,

$$\begin{gathered} \mathrm{p}\left(\mathrm{y}\ge 1\right)=\frac{9}{100} \\ =0.09 \end{gathered}$$

(b) Calculating the probability that exactly one taxpayer with incomes under $1 million will be audited

Here, the number of taxpayers is$\mathrm{n}=5$

The probability that taxpayers with income less than $1 million is p=0.01

Since,

$$\begin{gathered} q=1-p \\ =1-0.01 \\ =0.99 \end{gathered}$$

Also, if the random variable x follows a binomial distribution, then the probability function is defined as follows:

localid="1664383661850" $\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\left(\frac{n}{x}\right)\mathbf{}{\mathbf{\left(}\mathbf{p}\mathbf{\right)}}^{\mathbf{x}}\mathbf{}{\mathbf{\left(}\mathbf{q}\mathbf{\right)}}^{\mathbf{n}\mathbf{-}\mathbf{x}}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{(}\mathbf{x}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{1}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{n}\mathbf{)}$

Where,

P=probability of success on a single trial

n= Number of trials

x= Number of successes in n trials

q=1-p

Therefore,

The probability that strictly one will be audited by the IRS is given by,

$$\begin{gathered} \mathrm{p}(\mathrm{x}=1)=\left(\frac{5}{1}\right){(0.01)}^1{(0.99)}^{5-1}=\frac{5!}{1!{(5-1)}^1}{(0.01)}^1{(0.99)}^4 \\ =5\times 0.01\times 0.9606 \\ \mathrm{Therefore},\mathrm{p}(\mathrm{x}=1)=0.05 \end{gathered}$$

Also,

The probability that exactly one will be audited by the IRS is given by,

$$\begin{gathered} \mathrm{p}(\mathrm{x}>1)=1-\left[\mathrm{p}\right(\mathrm{x}=0)+\mathrm{p}(\mathrm{x}=1\left)\right] \\ =1-\left[\left(\frac{5}{0}\right){\left(0.01\right)}^0{\left(0.99\right)}^{5-0}+\left(\frac{5}{1}\right){\left(0.01\right)}^1{\left(0.99\right)}^{5-1}\right] \\ =1-\left[\frac{5!}{0!\left(5-0\right)!}{\left(0.01\right)}^0{\left(0.99\right)}^5+\frac{5!}{1!\left(5-1\right)!}{\left(0.01\right)}^1{\left(0.99\right)}^4\right] \\ =5\times 0.01\times 0.9606 \\ \mathrm{Therefore}, \\ \mathrm{P}(\mathrm{x}>1)=0.001 \end{gathered}$$

(c) Calculating the probability that exactly one taxpayer will be audited

The probability that a taxpayer with an income equal to $1 million or more is 0.09

Since,

$$\begin{gathered} \mathrm{q}=1-\mathrm{p} \\ =1-0.09 \\ =0.91 \end{gathered}$$

The probability that exactly one will be audited by the IRS is given by,

$$\begin{gathered} \mathrm{p}(\mathrm{y}=1)=\left(\frac{5}{1}\right){\left(0.09\right)}^1{\left(0.91\right)}^{5-1} \\ =\frac{5!}{1!\left(5-1\right)!}{\left(0.09\right)}^1{\left(0.91\right)}^4 \\ =0.3086 \end{gathered}$$

The probability the IRS will audit more than one is given by,

$$\begin{gathered} \mathrm{p}(\mathrm{y}>1)=1-\left[\mathrm{p}\right(\mathrm{y}=0)+\mathrm{p}(\mathrm{y}=1\left)\right] \\ =1-\left[\left(\frac{5}{0}\right){\left(0.09\right)}^0{\left(0.91\right)}^{5-0}+\left(\frac{5}{1}\right){\left(0.09\right)}^1{\left(0.91\right)}^{5-1}\right] \\ =1-\left[\frac{5!}{0!\left(5-0\right)!}{\left(0.09\right)}^0{\left(0.91\right)}^5+\frac{5!}{1!\left(5-1\right)!}{\left(0.09\right)}^1{\left(0.91\right)}^4\right] \\ \mathrm{P}(\mathrm{y}>1)=0.0674 \end{gathered}$$

(d) Calculating the probability that none of the taxpayers will be audited

Here, the number of taxpayers is $\mathrm{n}=2$

The probability that taxpayers with income less than $1 million is $\mathrm{p}=0.01$

The probability that none of these taxpayers will be audited by the IRS is given by,

role="math" localid="1664382856387" $$\begin{gathered} \mathrm{p}(\mathrm{x}=0)=\left(\frac{2}{0}\right){\left(0.01\right)}^0{\left(0.99\right)}^{2-0} \\ =\frac{2!}{0!\left(2-0\right)!}{\left(0.01\right)}^0{\left(0.99\right)}^2 \\ =0.9801 \end{gathered}$$

The probability that none of these taxpayers will be audited is by the IRS given by,

role="math" localid="1664382894860" $$\begin{gathered} \mathrm{p}(\mathrm{y}=0)=\left(\frac{2}{0}\right){\left(0.09\right)}^0{\left(0.91\right)}^{2-0} \\ =\frac{2!}{0!\left(2-0\right)!}{\left(0.09\right)}^0{\left(0.91\right)}^2 \\ =0.8281 \end{gathered}$$

The probability that none of these taxpayers will be audited by the IRS is given by,

$$\begin{gathered} \mathrm{p}\left(\mathrm{x}=0,\mathrm{y}=0\right)=\mathrm{p}\left(\mathrm{x}=0\right)\times \mathrm{p}\left(\mathrm{y}=0\right) \\ =0.9801\times 0.8281 \\ =0.8116 \end{gathered}$$

(e) Stating the assumptions

The critical assumptions to make in order to answer these questions are given below:

1. The events are independent.

2. The probability of success is the same in all the trials.