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

Answers

Answer 1

Complete Question

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 9-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.)

Answer:

The probability is  [tex]P(X > 2) = 0.0007 [/tex]

Step-by-step explanation:

From the question we are told that

The  probability of an IRS audit for a tax return with over $25,000 in income is p= 0.02

The sample size  n  =  9

Generally the distribution of IRS audit for a tax return follows a binomial distribution  

i.e  

         [tex]X  \~ \ \ \  B(n , p)[/tex]

and the probability distribution function for binomial  distribution is  

      [tex]P(X = x) =  ^{n}C_x *  p^x *  (1- p)^{n-x}[/tex]

Here C stands for combination hence we are going to be making use of the combination function in our calculators  

 

Generally the probability that a person is not audited at all is mathematically

     [tex]P(X = 0) =  ^{9}C_0 *  (0.02)^0 *  (1- 0.02 )^{9-0}[/tex]

      [tex]P(X = 0) =  ^{9}C_0 *  (0.02)^0 *  (0.98 )^{9-0}[/tex]

       [tex]P(X = 0) =  0.8337[/tex]

Generally the probability that a person is  audited once  is mathematically

           [tex]P(X = 1) =  ^{9}C_1 *  (0.02)^1 *  (1- 0.02 )^{9-1}[/tex]

           [tex]P(X = 1) =  9*  (0.02)^1 *  (0.98 )^{8}[/tex]

            [tex]P(X = 1) =  0.1531[/tex]

Generally the probability that a person is  audited twice is mathematically represented

           [tex]P(X = 2) =  ^{9}C_2 *  (0.02)^2 *  (1- 0.02 )^{9-2}[/tex]

           [tex]P(X = 2) =  36 *  (0.02)^2 *  (0.98 )^{7}[/tex]

            [tex]P(X = 2) =  0.0125 [/tex]

Generally the probability that a person is audited more than twice is mathematically represented

     [tex]P(X > 2) =  1 - [P(X = 0 ) + P(X = 1 )+ P(X =2)] [/tex]

=>    [tex]P(X > 2) =  1 - [0.8337 + 0.1531+0.0125 ] [/tex]

=>    [tex]P(X > 2) = 0.0007 [/tex]


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