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August 28, 2015
Posted by **Craig** on Saturday, June 6, 2009 at 3:13pm.

i don't know how to go about working this problem. i know its a discrete random variable problem and i know that i somehow have to work backwards, but i have NO idea how to set it up or work it. any and all guidance would be appreciated. thank you!!!

i don't expect you to work it for me, but i do need some help.

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**Craig**, Saturday, June 6, 2009 at 3:14pmwould this be a binomial-like should i consider it

applicable for study

not applicable for study

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**MathMate**, Sunday, June 7, 2009 at 9:46amYes, the binomial distribution is a discrete random variable problem.

Basically, if you screen N people, each having a success rate of 0.8. You would like to calculate the minimum candidates to have a minimum of 10 success with a probability of 0.939.

It is a very similar process to flipping coins, except the probability of success is 0.8 instead of 0.5.

The probability can be obtained by tabulation, but the a formula exists to calculate the probability of success of n times out of N candidates, with a probability of success of p.

P(n,N,p)=_{N}C_{n}*p^{N}*(1-p)^{(N-n)}

Let say, since the proability of success is 0.8, and 10 people are required, a good starting point is to screen 10/0.8=13 candidates.

So N=13, p=0.8,

P(10,13,0.8) = 0.246

P(11,13,0.8) = 0.268

P(12,13,0.8) = 0.179

P(13,13,0.8) = 0.055

Probability of retaining 10 or more candidates

= 0.246+0.268+0.179+0.055

= 0.747 < 0.939 required.

So you could screen 14 candidates and calculate the probability. Keep increasing the number of candidates until you get the probability of 0.939 or higher.

Total =

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**MathMate**, Sunday, June 7, 2009 at 9:46amOh yes, here's a link for more explanation:

http://stattrek.com/Lesson2/Binomial.aspx

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**MathMate**, Sunday, June 7, 2009 at 9:52amAlso, in case you are not familiar with the notation,

_{n}C_{r}is the combination function defined as:

_{n}C_{r}

= n!/((n-r)!r!)

=n(n-1)(n-2)...(n-r+1)/r!

For example,_{13}C_{10}

=13*12*11/(1*2*3)

=286

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**Craig**, Sunday, June 7, 2009 at 2:14pmthank you so much! i think i got it!

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**Jason**, Sunday, June 7, 2009 at 11:30pmwhat was the answer you finally got