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The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year. We are interested in the expected number of audits a person with that income has in a 16-year period. Assume each year is independent.

Find the probability that a person is audited more than twice. (Round your answer to four decimal places.)

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Answer:

The probability that a person is audited more than twice is 0.0037 .

Step-by-step explanation:

We are given that The chance of an IRS audit for a tax return with over $25,000 in income is about 2% per year.

We are interested in the expected number of audits a person with that income has in a 16-year period

Now we are supposed to find the probability that a person is audited more than twice.

So, We will use binomial distribution

So, Probability of success =p= 2% = 0.02

So, Probability of failure = q=1-0.02=0.98

n = 16

Now we are supposed to find the probability that a person is audited more than twice i.e. [tex]P(x \geq 2)[/tex]

So, [tex]P(x \geq 2)= 1-(P(x=0)+P(x=1)+P(x=2)[/tex]

Formula :[tex]^nC_x p^x q^{n-x}[/tex]

So,[tex]P(x \geq 2)= 1-(^{16}C_0 {(0.02)}^0 {(0.98)}^{16-0}+^{16}C_1 {(0.02)}^1 {(0.98)}^{16-1}+ ^{16}C_2 {(0.02)}^2 {(0.98)}^{16-2})[/tex][tex]P(x \geq 2)= 1-(\frac{16!}{0!(16-0)!} {(0.02)}^0 {(0.98)}^{16-0}+\frac{16!}{1!(16-1)!} {(0.02)}^1 {(0.98)}^{16-1}+\frac{16!}{2!(16-2)!} {(0.02)}^2 {(0.98)}^{16-2})[/tex]

[tex]P(x \geq 2)= 1-(0.9963)[/tex]

[tex]P(x \geq 2)= 0.0037[/tex]

Hence the probability that a person is audited more than twice is 0.0037 .

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