Question

The chances of a tax return being audited are about 16 in 1,000 if an income is less than $100,000 and 28 in 1,000 if an income is $100,000 or more. Complete partsa through e.a. What is the probability that a taxpayer with income less than $100,000 will be audited? With income of $100,000 or more?P(taxpayer with income less than $100,000 is audited) = 0.016(Type an integer or a decimal.)What is the probability that a taxpayer with income of $100,000 or more will be audited?P(taxpayer with income of $100,000 or higher is audited) = 0.028(Type an integer or a decimal.)b. If five taxpayers with incomes under $100,000 are randomly selected, what is the probability that exactly one will be audited? That more than one will be audited?P(x = 1) - (Round to four decimal places as needed.)S

Answer 1

The second part of the problem uses the concept of binomial probability. The formula to compute for the probability is

$P=_nC_x\cdot p^x\cdot(1-p)^{n-x}$

where nCx is the number of combinations that x will be selected from space n. The probability of success p for a single trial is computed already in part (a). 1-p represents the probability of failure.

For the probability that exactly one will be audited on five taxpayers, we have n = 5. The probability of success is computed on part (a) as 0.016. The value of nCx, in this case, is 5. We are now set to compute the value of probability for the first part. We have

$\begin{gathered} P(x=1)=5(0.016)^1(1-0.016)^{5-1} \\ P(x=1)=0.0750 \end{gathered}$

For the probability of x > 1, we will use the formula

$P(x>1)=1-\lbrack P(x=0)+P(x=1)\rbrack_{}$

where

$P(x=0)=\frac{n!}{x!(n-x)!}p^x(1-p)^{n-x}$

Using the value of n and p above at x = 0, we get

$\begin{gathered} P(x=0)=\frac{5!}{0!(5-0)!}(0.016)^0(1-0.016)^{5-0} \\ P(x=0)=0.9225 \end{gathered}$

The value of P(x = 1) is already computed above. Solving for P(x > 1), we get

$\begin{gathered} P(x>1)=1-\lbrack0.9225+0.0750\rbrack \\ P(x>1)=0.0025 \end{gathered}$

For part (c), we will use the same solving technique applied on part (b) but the probability that we will be using is p = 0.028 since we are dealing with the case for auditing with taxpayers with more than $100,000 income.

For P(x=1), we have

$\begin{gathered} P(x=1)=5(0.028)^1(1-0.028)^{5-1} \\ P(x=1)=0.1250_{} \end{gathered}$

For P(x>1), we have

$\begin{gathered} P(x=0)=\frac{5!}{0!(5-0)!}(0.028)^0(1-0.028)^{5-0} \\ P(x=0)=0.8676_{}_{} \\ P(x>1)=1-(0.8676+0.1250) \\ P(x>1)=0.0074 \end{gathered}$

For part (d), we will compute for the value of P(x = 0) for n = 2 for both the taxpayer with less than 100,000 income and the taxpayer with more than 100,000 income. we have

$\begin{gathered} P(x=0)=\frac{2!}{0!(2-0)!}(0.016)^0(1-0.016)^{2-0} \\ P(x=0)=0.9683 \end{gathered}$$\begin{gathered} P(x=0)=\frac{2!}{0!(2-0)!}(0.028)^0(1-0.028)^{2-0} \\ P(x=0)=0.9448_{}_{} \end{gathered}$

The product of these two probabilities will represent the final answer for the case that no taxpayer will be audited. We have

$P(none)=0.9683\cdot0.9448=0.9148$