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

04-01-2008, 06:36 PM
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:
x=greaterthan or equal to 3
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!

04-02-2008, 02:57 AM
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.

04-02-2008, 11:43 AM
Thanks, but I am not able to use the table, our teacher wants us to use the binomial formula.

04-02-2008, 11:55 AM
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.
I have seen this problem before.

04-02-2008, 11:55 AM
I did this with MathCad.
\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?

04-02-2008, 05:54 PM
Thanks for the feedback. Will do my arithmatic again, I don't want to have a minus 1 :D

04-26-2008, 07:15 AM
Hey, I was just doing some revision for my exams and wondered why the table thing gives a different answer.

Using this (http://cyber.gwc.cccd.edu/faculty/jmiller/Binom_Tab.pdf) 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?



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?