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A box contains 8 dark chocolates, 8 white chocolates, and 8 milk chocolates. I choose chocolates at random (yes, without replacement; I’m eating them). What is the chance that I have chosen 20 chocolates and still haven’t got all the dark ones? what is your answer to that
 one year ago
 one year ago
A box contains 8 dark chocolates, 8 white chocolates, and 8 milk chocolates. I choose chocolates at random (yes, without replacement; I’m eating them). What is the chance that I have chosen 20 chocolates and still haven’t got all the dark ones? what is your answer to that
 one year ago
 one year ago

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UsukiDollBest ResponseYou've already chosen the best response.0
choose milk chocolates choose white chocolates and just 4 dark chocolates XD
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
the question says that you still haven't got ALL the dark ones, so you COULD get 4 dark chocolates to 7 dark chocolates, but not the full 8
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
do you get the question now?
 one year ago

UsukiDollBest ResponseYou've already chosen the best response.0
I'm thinking milk white milk white milk white milk white milk white
 one year ago

perlBest ResponseYou've already chosen the best response.1
you wantthe probability of not getting all 8 dark chocolates. you are guaranteed at least 4 dark chocolates so the possibilities are 4 dark, 5 dark, 6 dark, 7 dark
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
yep, so if you find the probability of getting 8 dark chocolates, you can minus it from 1 and you'll get the probability of getting 4 dark, ... 7 dark
 one year ago

perlBest ResponseYou've already chosen the best response.1
there are 4 favorable cases, out of 5 total possible cases favorable cases : : exactly 4 dark , 16 nondark : exactly 5 dark 15 nondark exactly 6 dark 14 nondark exactly 7 dark 13 nondark total cases : exactly 4 dark , 16 nondark : exactly 5 dark 15 nondark exactly 6 dark 14 nondark exactly 7 dark 13 nondark exactly 8 dark , 12 nondark I see no reason not to treat these choices as equally likely. therefore the probability of not picking all dark ones is 4/5
 one year ago

perlBest ResponseYou've already chosen the best response.1
since they must add up to 20 , these are the only possibilites
 one year ago

perlBest ResponseYou've already chosen the best response.1
notice 3 dark 17 nondark is impossible
 one year ago

perlBest ResponseYou've already chosen the best response.1
the situation is constrained by two facts there must be no more than 16 nondark chocolates there must be at least 4 dark chocalates
 one year ago

perlBest ResponseYou've already chosen the best response.1
that produces (16,4) (15, 5 ) (14,6) (13, 7) (12,8)
 one year ago

perlBest ResponseYou've already chosen the best response.1
the situation is constrained by two facts you cannot have more than 16 nondark chocolates you cannot have more than 8 dark chocolates there must be at least 4 dark chocalates so the possibilities are (# nondark, # dark) (16,4) (15, 5 ) (14,6) (13, 7) (12,8) there is 4/5 favorable
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
the explanation seems reasonable enough
 one year ago

perlBest ResponseYou've already chosen the best response.1
i think its more complicated, let X = number of dark ones , Y = number of nondark chocolate how many ways can you choose 16 non dark and 4 dark? how many ways can you choose 15 nondark and 5 dark?
 one year ago

perlBest ResponseYou've already chosen the best response.1
P( chance not getting all dark in 20 draws) = P( X = 4 & Y= 16 or X= 5 & Y=15 or X=6 & Y=14 or X=7 or Y=13 )
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
that's what i thought too, but if you add up all the ways to not get all 8 dark chocolate, i'm like 90% sure you'll get 4/5...cause there are like 60 ways if you account for each individual selection
 one year ago

perlBest ResponseYou've already chosen the best response.1
P( X = 4 & Y= 16 or X= 5 & Y=15 or X=6 & Y=14 or X=7 or Y=13 ) = P ( X=4 & Y=16) + P(X=5 & Y=15) + ...
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
btw, what level of maths is this?
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
i meant like are you still in school or university?
 one year ago

perlBest ResponseYou've already chosen the best response.1
P(Y=16 & X=4) = P(Y=16) * P (X=4  Y = 16) =
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
oh, then you might not want to listen to me, i'm in yr 11 doing yr 12 maths so yea...
 one year ago

perlBest ResponseYou've already chosen the best response.1
i find this problem confusing
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
whoops, should have told you that at the start!
 one year ago

perlBest ResponseYou've already chosen the best response.1
im trying to force these into mutually exclusive possibilities
 one year ago

perlBest ResponseYou've already chosen the best response.1
you cannot have more than 16 nondark chocolates you cannot have more than 8 dark chocolates there must be at least 4 dark chocalates so the possibilities are (# nondark, # dark) (16,4) (15, 5 ) (14,6) (13, 7) (12,8)
 one year ago

perlBest ResponseYou've already chosen the best response.1
but are these equally likely?
 one year ago

perlBest ResponseYou've already chosen the best response.1
if these possibilities are equally likely to occur (and they are mutually exclusive), then the probability is 4/5
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
i'd assume so, unless some were heavier or bigger than the others (which isn't so)
 one year ago

perlBest ResponseYou've already chosen the best response.1
im use to working on problems like 5/12*4/11 , those sorts of problems
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
me too, I guess you could do that, but you'd be there forever!
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
I found the probability of getting all 8 chocolates, kinda took a while and subtracted it from 1
 one year ago

perlBest ResponseYou've already chosen the best response.1
can you show me your work
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
hmm..i guess you could use combinatorics
 one year ago

perlBest ResponseYou've already chosen the best response.1
because i want Probability ( 4 dark& 16 nondark) exactly
 one year ago

perlBest ResponseYou've already chosen the best response.1
\[\frac{ 8C4*16C16 }{24C20 }+\frac{ 8C5*16C15 }{24C20}+\frac{ 8C6*16C14}{24C20}+\frac{ 8C7*16C13 }{24C20 }\]
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
what do you get from that?
 one year ago

perlBest ResponseYou've already chosen the best response.1
the first is the case 4 dark & 16 nondark then 5 dark& 15 nondark
 one year ago

perlBest ResponseYou've already chosen the best response.1
the order of eating them does not count, but we should label in our mind the chocolates. if you want, label the dark chocolates 18, and label the nondark chocolates 924
 one year ago

perlBest ResponseYou've already chosen the best response.1
so now we have the problem dark chocolate = { 1,2,3,4,5,6,7,8} , nondark { 9,10,11,12,... 23,24} , the numbers are going to be labels for the chocolates, ok ?
 one year ago

perlBest ResponseYou've already chosen the best response.1
i could have used letters,
 one year ago

perlBest ResponseYou've already chosen the best response.1
so for instance, suppose we want four dark chocolates, and 16 nondark chocolate. so we do probability = # favorable outcomes / # total possibilities the favorable, first choose 4 dark chocolates which has a total of 8 choose 4 ways to do it, then multiply by that how many ways can you choose 16 nondark from 16 non dark. there is only one way. so im using multiplication rule (and probability)
 one year ago

perlBest ResponseYou've already chosen the best response.1
the denominator is the total number of ways to choose 20 chocolates , and there are 24 choose 20 ways to choose 20 chocolates
 one year ago

perlBest ResponseYou've already chosen the best response.1
anyways, show me your work, and we can compare answers . sorry if my explanation is confusing, im not exactly sure of the correct jargon
 one year ago

perlBest ResponseYou've already chosen the best response.1
so I got a probability of 629/759 = .8287
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
I tried using permutations so like: \[\frac{ nPr(16,16)*nPr(8,4)+nPr(16,15)*nPr(8,5)...etc }{ nPr(24,20) }\] but then the amswer was: \[\frac{ 5 }{ 245157 }\] which doesn't seem right...AT ALL... so i replaced all the P with C to use combinatorics and i got 629/759, which seems more accurate than permutations
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
it's close to 4/5 at least
 one year ago

perlBest ResponseYou've already chosen the best response.1
thats right, and yes its close to 4/5, but 4/5 would give you a wrong answer. i know how teachers are, lol
 one year ago

KABRICBest ResponseYou've already chosen the best response.0
perl your answer is right thx
 one year ago

perlBest ResponseYou've already chosen the best response.1
and you could solve it using the complement approach .
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
yea, basically what i was thinking
 one year ago

perlBest ResponseYou've already chosen the best response.1
chelsea, how did you get nCr using pretty print?
 one year ago

perlBest ResponseYou've already chosen the best response.1
\[1\frac{ \left(\begin{matrix}8 \\ 8\end{matrix}\right)*\left(\begin{matrix}16 \\ 12\end{matrix}\right) }{ \left(\begin{matrix}24 \\ 20\end{matrix}\right) }\]
 one year ago

perlBest ResponseYou've already chosen the best response.1
do you how it comes out the same ?
 one year ago

Chelsea04Best ResponseYou've already chosen the best response.0
you sort of lost me there
 one year ago

perlBest ResponseYou've already chosen the best response.1
P( X>7 ) = P(X=8 or X=9 or .. ) but X=9 and higher is impossible so we only need to look at X = 8
 one year ago

perlBest ResponseYou've already chosen the best response.1
we cant have 9 dark ones, since there arent 9 dark ones
 one year ago

perlBest ResponseYou've already chosen the best response.1
ok lets not pursue this further, i have a new question answering, follow me
 one year ago

WalterDeBest ResponseYou've already chosen the best response.0
Correct answer => http://629a3486.linkbucks.com
 11 months ago
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