anonymous
  • anonymous
among 18 computers in some store ,six have defects .five computers are randomly selected.compute the probability that all five computers are non defective? my solution: 12/18X11/17X10/16X9/15X8/14 .is it right?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
let me try something, just for practice :) \[P(x)=\frac{\binom{Number.fail}{0fail}\binom{number.success}{f.success}}{\binom{Total}{number.choosen}}\]
amistre64
  • amistre64
6C5 / 18C5
amistre64
  • amistre64
this tiny little screen, i might have got my success and failures sawpped up

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amistre64
  • amistre64
Total 18 6 are defective 5 chosen 5 not defective
amistre64
  • amistre64
\[P(x)=\frac{\binom{6}{0}\binom{12}{5}}{\binom{18}{5}}\]
anonymous
  • anonymous
i don understand ur solution :S :(
amistre64
  • amistre64
its the hyper geometric distribution. but lets see if we end up the same in the end 12.11.10.9.8 ------------ 18.17.16.15.14
amistre64
  • amistre64
yep, we are the same in the end
anonymous
  • anonymous
@spd two different methods to arrive at the same answer, your method in the question is correct. first is not times second is not given first is not times ... this will work amistre wrote a different method: total number of ways to choose 5 from 18 is \(\binom{18}{5}\) and you have to select all 5 from the 12 that are not defective, number of ways to do that is \(\binom{12}{5}\) so take the ratio \[\frac{\dbinom{12}{5}}{\dbinom{18}{5}}\] you will get the same answer

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