anonymous
  • anonymous
I worked the probability problem I just need someone to check the answer please! In a batch of 50 batteries, four are defective. If we select 3 batteries, what is the probability that at least 2 are good.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
I did P=1-P(0 good)P(1good) my answer is 99%
amistre64
  • amistre64
4 out of 50 are bad, so 46 out of 50 are good can we do a binomial approach?
amistre64
  • amistre64
\[\binom{50}{2}(\frac{46}{50})^2(\frac{4}{50})^{48}+\binom{50}{3}(\frac{46}{50})^{3}(\frac{4}{50})^{47}\]

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More answers

amistre64
  • amistre64
i think 50s are 3 in the binomial code ...
anonymous
  • anonymous
a binomial approach is used when looking for the probability of an exact number of things. i used the multiplication rule( that is what is used in the book to find at least prob.)
amistre64
  • amistre64
\[\binom{3}{2}(\frac{46}{50})^2(\frac{4}{50})^{1}+\binom{3}{3}(\frac{46}{50})^{3}(\frac{4}{50})^{0 }\]
amistre64
  • amistre64
exactly 2 or exactly 3 are at least 2
anonymous
  • anonymous
P(at least 2 good)=1-69/4900*1/4900 =.999999 =99%
anonymous
  • anonymous
P( 0 good) = ((4c3)(46c0))/(50c3) =1/4900
amistre64
  • amistre64
it is a high number ... im getting 98.2 or so
anonymous
  • anonymous
P(1 good)= ((4c2)(46c1))/(50c3) =69/4900
amistre64
  • amistre64
shold we assume replacement or not? ggb = (4*46*47)/(50*49*48) gbg = (4*46*47)/(50*49*48) bgg = (4*46*47)/(50*49*48) ggg = (46*47*48)/(50*49*48)
anonymous
  • anonymous
no replacement
amistre64
  • amistre64
3*(4*46*47)/(50*49*48)+(46*45*44)/(50*49*48) without replacement gets us about 99.51
anonymous
  • anonymous
So in other words 99% would be a correct answer
amistre64
  • amistre64
im not the one grading it, but it seems fair enough :)
anonymous
  • anonymous
Thanks!
amistre64
  • amistre64
youre welcome

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