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perl
 3 years 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
perl
 3 years 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

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UsukiDoll
 3 years ago
Best ResponseYou've already chosen the best response.0choose milk chocolates choose white chocolates and just 4 dark chocolates XD

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0do you get the question now?

UsukiDoll
 3 years ago
Best ResponseYou've already chosen the best response.0I'm thinking milk white milk white milk white milk white milk white

perl
 3 years ago
Best ResponseYou've already chosen the best response.1you 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yep, 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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1there 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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1since they must add up to 20 , these are the only possibilites

perl
 3 years ago
Best ResponseYou've already chosen the best response.1notice 3 dark 17 nondark is impossible

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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1that produces (16,4) (15, 5 ) (14,6) (13, 7) (12,8)

perl
 3 years ago
Best ResponseYou've already chosen the best response.1the 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the explanation seems reasonable enough

perl
 3 years ago
Best ResponseYou've already chosen the best response.1i 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?

perl
 3 years ago
Best ResponseYou've already chosen the best response.1P( 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 )

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0that'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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1P( 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) + ...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0btw, what level of maths is this?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i meant like are you still in school or university?

perl
 3 years ago
Best ResponseYou've already chosen the best response.1P(Y=16 & X=4) = P(Y=16) * P (X=4  Y = 16) =

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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1i find this problem confusing

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0whoops, should have told you that at the start!

perl
 3 years ago
Best ResponseYou've already chosen the best response.1im trying to force these into mutually exclusive possibilities

perl
 3 years ago
Best ResponseYou've already chosen the best response.1you 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)

perl
 3 years ago
Best ResponseYou've already chosen the best response.1but are these equally likely?

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

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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0me too, I guess you could do that, but you'd be there forever!

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0hmm..i guess you could use combinatorics

perl
 3 years ago
Best ResponseYou've already chosen the best response.1because i want Probability ( 4 dark& 16 nondark) exactly

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0what do you get from that?

perl
 3 years ago
Best ResponseYou've already chosen the best response.1the first is the case 4 dark & 16 nondark then 5 dark& 15 nondark

perl
 3 years ago
Best ResponseYou've already chosen the best response.1the 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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1so 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 ?

perl
 3 years ago
Best ResponseYou've already chosen the best response.1i could have used letters,

perl
 3 years ago
Best ResponseYou've already chosen the best response.1so 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)

perl
 3 years ago
Best ResponseYou've already chosen the best response.1the denominator is the total number of ways to choose 20 chocolates , and there are 24 choose 20 ways to choose 20 chocolates

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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1so I got a probability of 629/759 = .8287

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it's close to 4/5 at least

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0perl your answer is right thx

perl
 3 years ago
Best ResponseYou've already chosen the best response.1and you could solve it using the complement approach .

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yea, basically what i was thinking

perl
 3 years ago
Best ResponseYou've already chosen the best response.1chelsea, how did you get nCr using pretty print?

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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1do you how it comes out the same ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you sort of lost me there

perl
 3 years ago
Best ResponseYou've already chosen the best response.1P( 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

perl
 3 years ago
Best ResponseYou've already chosen the best response.1we cant have 9 dark ones, since there arent 9 dark ones

perl
 3 years ago
Best ResponseYou've already chosen the best response.1ok lets not pursue this further, i have a new question answering, follow me

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Correct answer => http://629a3486.linkbucks.com
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