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Likelihood of a tax return audit. At the beginning of each year, the Internal Revenue Service (IRS) releases information on the likelihood of a tax return being audited. In 2013, the IRS audited 1,242,479 individual tax returns from 145,236,429 filed returns; also, the IRS audited 25,905 returns from the total of 1,924,887 corporation returns filed (IRS 2014 Data Book).

a. Suppose an individual tax return is randomly selected. What is the probability that the IRS audited the return?

b. Refer to part a. Determine the probability that the IRS did not audit an individual return.

c. Suppose a corporation tax return is randomly selected. What is the probability that the IRS audited the return?

d. Refer to part c. Determine the probability that the IRS did not audit a corporation's return.

Step by step solution

01

Step-by-Step SolutionStep 1: Introduction

The probability is used to determine the likelihood of an event occurring. To clarify your position, probability refers to the possibility of an occurrence occurring.

$\text{Probability=}\frac{\text{Favourable outcome}}{\text{Total outcome}}$

02

Find the probability of individual tax return was audited

$\begin{array}{c}\text{P(individual return was audited)=}\frac{\text{12,42,479}}{\text{14,68,61,217}}\\ \text{=0.0088}\end{array}$

Hence, the required probability is0.0088.

03

Find the probability that the individual tax return was not audited

$\begin{array}{c}\text{P(individual return was not audited) = 1}鈭�\text{P(individual return was audited)}\\ \text{= 1}鈭�\text{0.0088}\\ \text{= 0.9912}\end{array}$

Hence, the required probability is0.9912.

04

Find the probability of corporation tax return was audited

$\begin{array}{c}\text{P(corporation return was audited)=}\frac{\text{25,905}}{\text{19,24,887}}\\ \text{=0.0134}\end{array}$

Hence, the required probability is0.0134.

05

Find the probability of corporation tax return was not audited

$\begin{array}{c}\text{P(corporation return was not audited) = 1}鈭�\text{P(corporation return was audited)}\\ \text{= 1}鈭�\text{0.0134}\\ \text{= 0.9866}\end{array}$

Hence, the required probability is0.9866.

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