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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?
 one year ago
 one year 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?
 one year ago
 one year ago

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germanphysicsBest ResponseYou've already chosen the best response.0
*that at least 17 succed.
 one year ago

KingBest ResponseYou've already chosen the best response.1
probability that 17 students succeed is 85%
 one year ago

rebeccaskell94Best ResponseYou've already chosen the best response.1
lol are you *in* a test? x)
 one year ago

KingBest ResponseYou've already chosen the best response.1
but since each of their's chance is NOT 100% then .....??
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
i think it's 120c3 * 0.25*0.25*0.25
 one year ago

germanphysicsBest ResponseYou've already chosen the best response.0
Nope, but I'm going to be in at least 2 hours :P
 one year ago

rebeccaskell94Best ResponseYou've already chosen the best response.1
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 ;)
 one year ago

germanphysicsBest ResponseYou've already chosen the best response.0
So 120c3*0,25*0,25*0,25 = 16,8125 % ?
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
oh.....i was 16,8125 ?!
 one year ago

germanphysicsBest ResponseYou've already chosen the best response.0
dw:1340020008729:dw
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
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
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
@satellite73 m i correct?
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

germanphysicsBest ResponseYou've already chosen the best response.0
Yeah. Same! Thank you very much!
 one year ago
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