probability tax payers are randomly sel.. please help

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Sonja

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Don't know where to start.

According to the CRA, the chances of your tax return being audited are 3 in 100 if your income is less than 60,000 and 8 in 100 if your income is more than 60,000

a) if seven taxpayers with incomes over 60,000 are randomly selected, what is the probability that at least two of them will be audited?

b) same question but under 60,000 are randomly selected

Please, please help my tutor will not return any phone calls.. it's an on-line course.
 
a)
probability of a audit,P=.08
then Q=.97

the sum of P+Q is 1
1=[P+Q]

if 7 taxpayers are chosen the possible outcomes are
1=[P+Q]^7
1=P^7+7P^6Q^1 + 21P^5Q^2+35P^4Q^3+35P^3Q^4+21P^2Q^5+7P^1Q^6+1Q^7

at least two is the sum of the P^2+P^3+P^4+P^5+P^6 +P^7 th terms,

but this is also the same as 1- the P^1+P^2 term
1-[P^7+7P^6Q] answer


b)
let P=.03
then Q=.97

at least 2 are chosen is the same as not choosing 1 or 2 taxpayers
1-P^7 - 7P^6Q answer

Arthur
 
Sonja said:
Don't know where to start.

According to the CRA, the chances of your tax return being audited are 3 in 100 if your income is less than 60,000 and 8 in 100 if your income is more than 60,000

a) if seven taxpayers with incomes over 60,000 are randomly selected, what is the probability that at least two of them will be audited?
Click to expand...

This is a binomial probability. The probability of earning over $60,000 and getting audited is 8% or p=0.08, q=1-0.08=0.92

\(\displaystyle \L\\\sum_{k=2}^{7}C(7,k)p^{k}q^{7-k}\)

Since it says, at least 2, the sum means you must use the case for 2,3,4,5,6,7 and add them up.

[quote:34je7t4q]b) same question but under 60,000 are randomly selected
Click to expand...

Same principal as above except p=0.03

Please, please help my tutor will not return any phone calls.. it's an on-line course.[/quote:34je7t4q]
 
so i have to do the ork for x= 2,3,4,5,6,7 and add those up
is this the formula i am using
n!
____Px (1-p) n-x
x! (n-x)!
 
It's kind of rough to read your post, but if this is what you mean:

\(\displaystyle \L\\\frac{n!}{(n-x)!x!}p^{x}(1-p)^{n-x}\)

Then yes. Note, I used a k instead of an x in my previuos formula. It's the same thing.
 
yes that is what i was trying to write lol sorry!!

im not quite sure though which numbers to plug in after i use 2,3,4,5,6,7 for x when doing all the seperate formulas.
So i will have to erite the formula 6 times and plug the different values of x in right
 
Do you have a calculator which will do sums?.

Here's for x=2:

\(\displaystyle \L\\C(7,2)(.08)^{2}(0.92)^{5}=0.08858...\)

What this is saying is, the probabiltiy of exactly 2 people being audited is about 9%. You need at least 2. Which is all of them from 2 to 7.

Note, \(\displaystyle \L\\C(n,x)=\frac{n!}{(n-x)!x!}\)

Now do the same thing for 3 through 7 and sum them up. A calculator

with \(\displaystyle \sum\) would be nice. But if not, add them up separately.
 
i have a scientific calculator but it doesn't have that button you showed up above. i did the first question a) and my answer after i added everything was approx:
.63166 ? that at least 2 tax payers will be audited.. i don't know??lol

i don't know how to start B) either Im in a real jam.. im definetely not a math girl lol
 
Math, like everything else, takes practice. Let me give you an answer to shoot for.

The proabability of getting audited is 8% if you earn over 60 grand. The

probability at least 2 people, out of 7 randomly chosen, getting audited is

\(\displaystyle \L\\\sum_{x=2}^{7}C(7,x)(.08)^{x}(.92)^{7-x}\approx{0.102}\)

About 10.2%

\(\displaystyle \L\\\underbrace{C(7,2)(.08)^{2}(.92)^{5}}_{\text{21p^2q^5}}+\underbrace{C(7,3)(.08)^{3}(.92)^{4}}_{\text{35p^3q^4}}\\+\underbrace{C(7,4)(.08)^{4}(.92)^{3}}_{\text{35p^4q^3}}+\underbrace{C(7,5)(.08)^{5}(.92)^{2}}_{\text{21p^5q^2}}\\+\underbrace{C(7,6)(.08)^{6}(.92)^{1}}_{\text{7p^6q}}+\underbrace{C(7,7)(.08)^{7}(.92)^{0}}_{\text{p^7}}\)

See the pattern?.

Think of expanding:

\(\displaystyle \L\\(p+q)^{7}=\underbrace{p^{7}+7p^{6}q+21p^{5}q^{2}+35p^{4}q^{3}+35p^{3}q^{4}+21p^{2}q^{5}}_{\text{what you use}}+7pq^{6}+q^{7}\)

You are using "at least 2". So, you use all the terms from p^2 to p^7.
This is the same thing as what we used above.

If it would've asked for "exactly 2", then you would just use 21p^2q^5.

Starting to see?. The binomial is very mechanical and patterned. Once you see it, it's easy to use.

Also, look up Pascal's triangle. It's can be big help.
 
my brain is is jumbles..
i got the same answer you gave me for x=2 so i figured that. I did all the others up to 7 in the same form and still get the .8 blah blah thing

so all look at what you gave me and go through it again.
for b. if its the probability of none of them getting audited what formula do i start with or i should say in which manner. do i do from 0 to 7 and add them all together?
Im so sorry for taking up so much time of yours. if you lived in Ontario I would just get you to tutor me lol
 
As I said, part b seems to be identical to part a, except use p=0.03 and q=0.97. part b doesn't say choosing none. It says, "Of 7 randomly selected who make under $60,000, what's the probability at least 2 will be audited". The chances for someone earning < $60,000 being audited is 3%.
Go from there.
 
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