anonymous
  • anonymous
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
Sorry for the usurping of my humor and all
alekos
  • alekos
don't worry about it. lets get back to the question
anonymous
  • anonymous
Yeah

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anonymous
  • anonymous
So I thought the answer would be the reasonable 4/5*4/5*4/5
anonymous
  • anonymous
Because I ace the 3 exams out of 4
alekos
  • alekos
probability (3 out 4 tests with >90%) = P(1st >90%) x P(2nd >90%) x P(3rd >90%) x P(4th <90%)
alekos
  • alekos
P(any test >90%) = 4/5
anonymous
  • anonymous
Looks like a permutation
alekos
  • alekos
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?
anonymous
  • anonymous
Every time it's 1/5
anonymous
  • anonymous
Oh so 4/5*4/5*4/5*1/5?
alekos
  • alekos
yep thats it!! so whats the final probability?
anonymous
  • anonymous
Like when they say every superman falls from a tree
anonymous
  • anonymous
64/600
anonymous
  • anonymous
with GCF, 32/300=16/150
anonymous
  • anonymous
8/75 yes?
alekos
  • alekos
just wait one minute. i'm re-thinking my answer
anonymous
  • anonymous
ok
alekos
  • alekos
its not quite right There are 4 ways that this can happen so we have to multiply the final probability by 4
alekos
  • alekos
so we have 4 x (4x4x4)/(5x5x5x5)
alekos
  • alekos
256/625
alekos
  • alekos
because its 3 out of 4, we could have 4 different ways that this can happen
alekos
  • alekos
battery has run out on my notebook so i gotta go
mathmate
  • mathmate
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!
dan815
  • dan815
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alekos
  • alekos
@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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