Mathematics, 26.10.2020 16:40 Jasten

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

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Answer from: chaundreascott743

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.)

The probability is P(X 2) = 0.0007

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

$X \~ \ \ \ B(n , p)$

and the probability distribution function for binomial distribution is

$P(X = x) = ^{n}C_x * p^x * (1- p)^{n-x}$

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

$P(X = 0) = ^{9}C_0 * (0.02)^0 * (1- 0.02 )^{9-0}$

$P(X = 0) = ^{9}C_0 * (0.02)^0 * (0.98 )^{9-0}$

P(X = 0) = 0.8337

Generally the probability that a person is audited once is mathematically

$P(X = 1) = ^{9}C_1 * (0.02)^1 * (1- 0.02 )^{9-1}$

$P(X = 1) = 9* (0.02)^1 * (0.98 )^{8}$

P(X = 1) = 0.1531

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

$P(X = 2) = ^{9}C_2 * (0.02)^2 * (1- 0.02 )^{9-2}$

$P(X = 2) = 36 * (0.02)^2 * (0.98 )^{7}$

P(X = 2) = 0.0125

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

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

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

=> P(X 2) = 0.0007

Answer from: Quest

coin flip

non-answer:

choosing

sorry if this isnt what you're looking for i dont really understand the question

Answer from: Quest

the simplified form is

$\frac{x^2(2x+3)}{2(4x-3)}$

step-by-step explanation:

we want to simplify the expression,

$\frac{\frac{6x+9}{15x^2} }{\frac{16x-12}{10x^4} }$

let us change the middle bar to a normal division sign to obtain,

$\frac{6x+9}{15x^2}\div\frac{16x-12}{10x^4}$

we multiply the first fraction by the reciprocal of the second fraction to obtain,

$\frac{6x+9}{15x^2}\times \frac{10x^4}{16x-12}$

we factor to obtain,

$\frac{3(2x+3)}{3\times5\times x^2}\times \frac{2\times5\times x^2\times x^2}{4(4x-3)}$

we cancel the common factors to get,

$\frac{(2x+3)}{1}\times \frac{ x^2}{2(4x-3)}$

we simplify to get,

$\frac{x^2(2x+3)}{2(4x-3)}$

the correct answer is b

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