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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
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry for the usurping of my humor and all

alekos
 one year ago
Best ResponseYou've already chosen the best response.1don't worry about it. lets get back to the question

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I thought the answer would be the reasonable 4/5*4/5*4/5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because I ace the 3 exams out of 4

alekos
 one year ago
Best ResponseYou've already chosen the best response.1probability (3 out 4 tests with >90%) = P(1st >90%) x P(2nd >90%) x P(3rd >90%) x P(4th <90%)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Looks like a permutation

alekos
 one year ago
Best ResponseYou've already chosen the best response.1not 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
 one year ago
Best ResponseYou've already chosen the best response.0Oh so 4/5*4/5*4/5*1/5?

alekos
 one year ago
Best ResponseYou've already chosen the best response.1yep thats it!! so whats the final probability?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Like when they say every superman falls from a tree

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0with GCF, 32/300=16/150

alekos
 one year ago
Best ResponseYou've already chosen the best response.1just wait one minute. i'm rethinking my answer

alekos
 one year ago
Best ResponseYou've already chosen the best response.1its not quite right There are 4 ways that this can happen so we have to multiply the final probability by 4

alekos
 one year ago
Best ResponseYou've already chosen the best response.1so we have 4 x (4x4x4)/(5x5x5x5)

alekos
 one year ago
Best ResponseYou've already chosen the best response.1because its 3 out of 4, we could have 4 different ways that this can happen

alekos
 one year ago
Best ResponseYou've already chosen the best response.1battery has run out on my notebook so i gotta go

mathmate
 one year ago
Best ResponseYou've already chosen the best response.1You 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 \(\large C(n,s) p^s (1p)^{ns}\) and C(n,s) is combination, n choose s, equal to n!/(s!(ns)!) and n=4, s=3. P(3 out of 4) = \(C(4,3)~ 0.8^3 (10.8)^{43}\) Take out your calculator and evaluate the probability accordingly!

alekos
 one year ago
Best ResponseYou've already chosen the best response.1@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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