anonymous
  • anonymous
***
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
binomial for this one \[n=5,k=4,p=\frac{2}{3},1-p=\frac{1}{3}\]
anonymous
  • anonymous
there are 5 combinations of four days... do i just raise (2/3) to the fourth power?
anonymous
  • anonymous
use \[P(x=4)=\binom{5}{4}\left(\frac{2}{3}\right)^4\left(\frac{1}{3}\right)\]

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anonymous
  • anonymous
ohhhh i see
anonymous
  • anonymous
yeah 5 combinations of 4 days, and \(\frac{2}{3}\) gets raised to the power of 4,
anonymous
  • anonymous
but also one failure, so need the \(\frac{1}{3}\) there as well
anonymous
  • anonymous
got it. Thanks so much!
anonymous
  • anonymous
yw

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