anonymous
  • anonymous
two out of six computers in a lab have problems with hard drives .if three computers are selected at random for inspection,what is the probability that none of them has hard drive problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
\[\frac{2}{6}=\frac{1}{3}\] is the probability that one has a hard drive problem, so probability that a computer does not have one is \(1-\frac{1}{3}=\frac{2}{3}\)
anonymous
  • anonymous
then the probability all three selected do not have a problem is \((\frac{2}{3})^3=\frac{8}{27}\)
anonymous
  • anonymous
oh wow sorry that was way wrong i did not read the "in the lab" part!!

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anonymous
  • anonymous
i thought there were many many computers, not just six it should be \[\frac{4}{6}\times \frac{3}{5}\times \frac{2}{4}\] as your answer
anonymous
  • anonymous
ok but here is my solution plz tell me if m right. 2 are defective so 4 will be non defective P(non defective)=4/6X3/5X2/4X1/3=0.06 1- P(defective among selected)=2/3 X1/2=1/3=0.6 now P(non defective)=0.06X.6
experimentX
  • experimentX
must be 4/6x3/5x2/4
anonymous
  • anonymous
how is that? 4 would be good right?? plz explain it in steps
experimentX
  • experimentX
first let's see the question properly ... "what is the probability that none of them has hard drive problem" here the question asks that: 3 computers selected out of 6 has NOT hard drive problem. now we have to select 3 computers ..ie 3 places __X__X__ what is probability for 1st place that it does not have hard drive problem similarly for 2nd place and 3 rd place.
anonymous
  • anonymous
ok
anonymous
  • anonymous
3/6X2/5X1/4
experimentX
  • experimentX
NO .. for first place you have 4 non-damaged pc out of 6
anonymous
  • anonymous
ok for the first place its 4/6 for 2nd place it wud be 3/5 for 3rd 2/4 ok got it

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