anonymous
  • anonymous
A box of 15 parts contains 7 that are defective. A worker picks parts one at a time and attempts to install them. Find the probability of each outcome in​ (a) through​ (d). ​(a) The first two chosen are both good. nothing ​(Round to four decimal places as​ needed.)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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mathmate
  • mathmate
Please check your question, which probably contains typo(s) and incomplete. Cannot have 15 parts of which 77 are defective.
anonymous
  • anonymous
it a 7
anonymous
  • anonymous
P(two are good) \[=\frac{C _{2}^{8} }{C_{2}^{15} }\]=?

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anonymous
  • anonymous
30
anonymous
  • anonymous
ok im so lost 8/15
anonymous
  • anonymous
and 054
anonymous
  • anonymous
pl help me out
anonymous
  • anonymous
im so lost and dum
mathmate
  • mathmate
(a) Initially, there are 8 good out of 15, so probability of getting the first good one is 8/15. For the next one, there are only 7 good out of 14, so probability is 7/14. For this two step experiment, the probability of getting 2 good out of two is therefore (8/15)*(7/14)=4/15. Alternatively, you could use the formula by @Surjithayer

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