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Mathematics 25 Online
OpenStudy (anonymous):

40 students live in a dormitory. 5 have swine flu. If 4 students are selected at random, what is the probability that 1 or 2 of the 4 have swine flu?

OpenStudy (anonymous):

(5/40)(4/39)((3/37)= 1.03950104*10^-3

OpenStudy (anonymous):

This is not the correct answer

OpenStudy (anonymous):

can u tell me what it is please, i suck at probablity, my weakest subject

OpenStudy (anonymous):

Google'd and discovered this.... http://www.jiskha.com/display.cgi?id=1298256153

OpenStudy (anonymous):

no problem. so I found (40 nCr 4)= 91390 then \[(35 nCr 3)(5 nCr 1)+(35 nCr 2)(5 nCr 2)=38675\] then \[38675\div91390=0.423\]

OpenStudy (anonymous):

I actually posted that crsf29 and they didnt have the right answer either

OpenStudy (anonymous):

Did it work out for you? I could dig out my ProbStat book and help you, but it would take me about 20 minutes and I'd wake my roomies...Sorry.

OpenStudy (anonymous):

you dont have to do that I figured it out

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