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anonymous

  • one year ago

So I can get by probability of 4/5 more than 90 percent in any quiz but I don't know how to calculate the probability by which I can get 3 out of 4 tests more 90

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  1. anonymous
    • one year ago
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    Sorry for the usurping of my humor and all

  2. alekos
    • one year ago
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    don't worry about it. lets get back to the question

  3. anonymous
    • one year ago
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    Yeah

  4. anonymous
    • one year ago
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    So I thought the answer would be the reasonable 4/5*4/5*4/5

  5. anonymous
    • one year ago
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    Because I ace the 3 exams out of 4

  6. alekos
    • one year ago
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    probability (3 out 4 tests with >90%) = P(1st >90%) x P(2nd >90%) x P(3rd >90%) x P(4th <90%)

  7. alekos
    • one year ago
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    P(any test >90%) = 4/5

  8. anonymous
    • one year ago
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    Looks like a permutation

  9. alekos
    • one year ago
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    not quite. its just a multiplication of probabilities because you're looking at getting 3 tests >90% and 1 test <90%, so you just multiply out. What do you think that the probability of getting <90% will be?

  10. anonymous
    • one year ago
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    Every time it's 1/5

  11. anonymous
    • one year ago
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    Oh so 4/5*4/5*4/5*1/5?

  12. alekos
    • one year ago
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    yep thats it!! so whats the final probability?

  13. anonymous
    • one year ago
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    Like when they say every superman falls from a tree

  14. anonymous
    • one year ago
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    64/600

  15. anonymous
    • one year ago
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    with GCF, 32/300=16/150

  16. anonymous
    • one year ago
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    8/75 yes?

  17. alekos
    • one year ago
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    just wait one minute. i'm re-thinking my answer

  18. anonymous
    • one year ago
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    ok

  19. alekos
    • one year ago
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    its not quite right There are 4 ways that this can happen so we have to multiply the final probability by 4

  20. alekos
    • one year ago
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    so we have 4 x (4x4x4)/(5x5x5x5)

  21. alekos
    • one year ago
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    256/625

  22. alekos
    • one year ago
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    because its 3 out of 4, we could have 4 different ways that this can happen

  23. alekos
    • one year ago
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    battery has run out on my notebook so i gotta go

  24. mathmate
    • one year ago
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    You have tests that are (assumed) independent, probability of success (\(\ge\)90%) is constant at p=0.8 throughout You have a number of Bernoulli trials (one of two outcomes) The number of trials is known (4). Under all these conditions, you can use the binomial distribution, where the number of success (s) is given by \(\large C(n,s) p^s (1-p)^{n-s}\) and C(n,s) is combination, n choose s, equal to n!/(s!(n-s)!) and n=4, s=3. P(3 out of 4) = \(C(4,3)~ 0.8^3 (1-0.8)^{4-3}\) Take out your calculator and evaluate the probability accordingly!

  25. dan815
    • one year ago
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    |dw:1437144554838:dw|

  26. alekos
    • one year ago
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    @mathmate, best to keep it simple for this guy, but your right. comes to the same answer I got, (4/5)^4

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