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

- anonymous

- schrodinger

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- anonymous

Sorry for the usurping of my humor and all

- alekos

don't worry about it.
lets get back to the question

- anonymous

Yeah

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## More answers

- anonymous

So I thought the answer would be the reasonable 4/5*4/5*4/5

- anonymous

Because I ace the 3 exams out of 4

- alekos

probability (3 out 4 tests with >90%) = P(1st >90%) x P(2nd >90%) x P(3rd >90%) x P(4th <90%)

- alekos

P(any test >90%) = 4/5

- anonymous

Looks like a permutation

- 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

Every time it's 1/5

- anonymous

Oh so 4/5*4/5*4/5*1/5?

- alekos

yep thats it!!
so whats the final probability?

- anonymous

Like when they say every superman falls from a tree

- anonymous

64/600

- anonymous

with GCF, 32/300=16/150

- anonymous

8/75 yes?

- alekos

just wait one minute. i'm re-thinking my answer

- anonymous

ok

- alekos

its not quite right
There are 4 ways that this can happen so we have to multiply the final probability by 4

- alekos

so we have 4 x (4x4x4)/(5x5x5x5)

- alekos

256/625

- alekos

because its 3 out of 4, we could have 4 different ways that this can happen

- alekos

battery has run out on my notebook so i gotta go

- 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

|dw:1437144554838:dw|

- 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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