Binomial formula: prob. that 3 of 5 returns are audited

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Please help!
I have tried this problem a few times, and don't feel confident about my answer.
Here's the Question:
A financial company prepares tax returns for individuals. According to IRS, individuals in a certain high income bracket are audited at a rate of 1.5%. The company prepares 5 tax returns for individuals in that high tax bracket. Find the probability that at least 3 are audited.

My work:
n=5
x=greaterthan or equal to 3
p=.015
q=.985
5C3 (.015)^3(.985)^2 + 5C4 (.015)^4(.985)^1 + 5C5 (.015)^5(.985)^0
= 10 (.00000328) + 5 (.0000000497) + 1 (.000000000759)
= .0000328 + .000000249 + .000000000759
= .0000329951

Basically, I don't feel confident in my answer. Any feedback is good feedback to me. Thanks!
 
Re: Binomial formula

I don't know if below is right or your answer, but this is the way I would do it. I seem to be at the same level as you.

If you're looking for the probability that x is greater than or equal to 3, I *think* you can do one take the probability of 2 or less.

To do this, I'd use a binomial table as it's quicker: http://cyber.gwc.cccd.edu/faculty/jmiller/Binom_Tab.pdf

n=5, p=0.15, x=2

Look up 2 or less in the corresponding table and subtract from 1.

Anyway, I don't know if that is right. If you need to use the formula I'd wait for confirmation from someone else.
 
Thanks, but I am not able to use the table, our teacher wants us to use the binomial formula.
 
Why don't you feel confident? Your expression for the binomial is correct. Make sure your arithmetic is correct, or you may lose a point.
8-)
I have seen this problem before.
 
I did this with MathCad.
\(\displaystyle \sum\limits_{k = 3}^5 {{\binom{5}{k}}\left( {.015} \right)^k \left( {.985} \right)^{5 - k} } = {\rm{0}}{\rm{.00003299518125}}\)
Does that make you feel better about it?
 
Thanks for the feedback. Will do my arithmatic again, I don't want to have a minus 1 :D
 
Hey, I was just doing some revision for my exams and wondered why the table thing gives a different answer.

Using this I looked up n=5, p=0.15, x=2 which was 0.1382.

That gave the probability of 2 or less, so to get 3 or more I did 1 - 0.1382 = 0.8618.

I'm sure I've done these types of questions like this in the past but was just wondering why my answer differed...

Where have I gone wrong?

Thanks.

EDIT:

I realised it's because I need p=0.015 (not 0.15) which the tables don't give :oops:. If I had this, would my method have worked?
 
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