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germanphysics

  • 3 years ago

Need fast help: 20 students are writing a test. Each of them has a 75% chance to succed. How high is the chance, that 17 students succed?

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  1. germanphysics
    • 3 years ago
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    *that at least 17 succed.

  2. King
    • 3 years ago
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    probability that 17 students succeed is 85%

  3. rebeccaskell94
    • 3 years ago
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    lol are you *in* a test? x)

  4. King
    • 3 years ago
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    but since each of their's chance is NOT 100% then .....??

  5. A.Avinash_Goutham
    • 3 years ago
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    i think it's 1-20c3 * 0.25*0.25*0.25

  6. germanphysics
    • 3 years ago
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    Nope, but I'm going to be in at least 2 hours :P

  7. rebeccaskell94
    • 3 years ago
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    haha good cramming ;P I was gonna try to answer and then decided I didn't want to make you fail x) Good luck though! oh and *succeed ;)

  8. germanphysics
    • 3 years ago
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    So 1-20c3*0,25*0,25*0,25 = 16,8125 % ?

  9. A.Avinash_Goutham
    • 3 years ago
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    oh.....i was 16,8125 ?!

  10. A.Avinash_Goutham
    • 3 years ago
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    *it

  11. germanphysics
    • 3 years ago
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    |dw:1340020008729:dw|

  12. A.Avinash_Goutham
    • 3 years ago
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    oh i l tell u my reasoning......... each one has a 75% probalilty of success...... so for 17 to scuceed........u need to select 17 students.......... multiply it with probabilty of success......ie 0.75^17..........or............3 fails ie 1- 20c3* 0.25^3

  13. A.Avinash_Goutham
    • 3 years ago
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    @satellite73 m i correct?

  14. anonymous
    • 3 years ago
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    this is going to take a bunch of computation "at least' 17 means 17 or 18 or 19 or 20 use binomial for each \(P(x=20)=(.75)^{20}\) \[p(x=19)=20\times .75^{19}\times .25\] \[P(x=18)=\dbinom{20}{2}.75^{18}\times .25^2\] \[P(x=17)=\dbinom{20}{3}.75^{17}\times .25^3\] add these up

  15. germanphysics
    • 3 years ago
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    Ah, alright!

  16. anonymous
    • 3 years ago
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    this is what i get, but check the calculations http://www.wolframalpha.com/input/?i=.75^20%2B20*.75^19*.25%2B190*.75^18*.25^2%2B1140*.75^17*.25^3

  17. germanphysics
    • 3 years ago
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    Yeah. Same! Thank you very much!

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