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]
