Question

The chances of a tax return being audited are about 14 in 1,000 if an income is less than​ $100,000 and 36 in 1,000 if an income is​ $100,000 or more. Complete parts a through e

B. If three tax payers with incomes under​ $100,000 are randomly​ selected, what is the probability that exactly one will be​ audited? That more than one will be​ audited?

p(x=1)=

​(Round to four decimal places as​ needed.)

What is the probability that more than one will be​ audited?

P(x>1)=

​(Round to four decimal places as​ needed.)

c. Repeat part b assuming that three taxpayers with incomes of​ $100,000 or more are randomly selected.

P(x=1)=

​(Round to four decimal places as​ needed.)

What is the probability that more than one will be​ audited?

P(x>1)=

​(Round to four decimal places as​ needed.)

d. If two taxpayers with incomes under​ $100,000 are randomly selected and two with incomes more than​ $100,000 are randomly​ selected, what is the probability that none of these taxpayers will be​ audited?

​P(none of the taxpayers will be audited)=

Answer 1

SOLUTION:

From given data,

The chances of a tax return being audited are about 14 in 1,000 if an income is less than​ $100,000 and 36 in 1,000 if an income is​ $100,000 or more.

P( income is less than​ $100,000) = 14/1000 = 0.014

P( income is​ $100,000 or more) = 36/1000 = 0.036

B. If three tax payers with incomes under​ $100,000 are randomly​ selected, what is the probability that exactly one will be​ audited? That more than one will be​ audited?

Let X = number of people with income less than $100000 who audited then

X binomial (n = 3 , p = 0.014)

p(x=1)= ³C₁ * (0.014)_¹  * (1-0.014)^(3-1)

^{= 3 * 0.014 * 0.972196}

^{= 0.04083}

What is the probability that more than one will be​ audited?

= 1-[ ³C₀ * (0.014)_⁰  * (1-0.014)^(3-0) +  0.04083 ]

= 1- [1*1*0.95858 + 0.04083]

= 1- 0.99941

= 0.00059

c. Repeat part b assuming that three taxpayers with incomes of​ $100,000 or more are randomly selected.

Let X = number of people with income less than $100000 who audited then

X binomial (n = 3 , p =  0.036)

p(x=1)= ³C₁ * ( 0.036)_¹  * (1- 0.036)^(3-1)

⁼ 3 * 0.036 * 0.929296

^{= 0.1003639}

What is the probability that more than one will be​ audited?

= 1-[ ³C₀ * (0.036)_⁰  * (1-0.036)^(3-0) +  0.1003639]

= 1- [1*1*0.8958413 + 0.1003639]]

= 1- 0.9962052

= 0.0037948

d. If two taxpayers with incomes under​ $100,000 are randomly selected and two with incomes more than​ $100,000 are randomly​ selected, what is the probability that none of these taxpayers will be​ audited?

X binomial (n = 2 , p = 0.014)

p(x=0)= ²C₀ * (0.014)_⁰ * (1-0.014)^(2-0)

= 1*1*0.972196

= 0.972196

X binomial (n = 2 , p = 0.036)

p(x=0)= ²C₀ * (0.036)_⁰ * (1-0.036)^(2-0)

= 1*1*0.929296

= 0.929296

P(none of the taxpayer will be audited) = 0.972196 *0.929296

P(none of the taxpayer will be audited) = 0.90345