perl
  • perl
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
dont really get it
UsukiDoll
  • UsukiDoll
choose milk chocolates choose white chocolates and just 4 dark chocolates XD
anonymous
  • anonymous
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

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perl
  • perl
oh woops
perl
  • perl
i thought it said 'any'
anonymous
  • anonymous
do you get the question now?
perl
  • perl
yes
UsukiDoll
  • UsukiDoll
I'm thinking milk white milk white milk white milk white milk white
perl
  • perl
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
anonymous
  • anonymous
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
perl
  • perl
there are 4 favorable cases, out of 5 total possible cases favorable cases : : exactly 4 dark , 16 non-dark : exactly 5 dark 15 non-dark exactly 6 dark 14 non-dark exactly 7 dark 13 non-dark total cases : exactly 4 dark , 16 non-dark : exactly 5 dark 15 non-dark exactly 6 dark 14 non-dark exactly 7 dark 13 non-dark exactly 8 dark , 12 non-dark 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
  • perl
since they must add up to 20 , these are the only possibilites
perl
  • perl
notice 3 dark 17 non-dark is impossible
perl
  • perl
the situation is constrained by two facts there must be no more than 16 non-dark chocolates there must be at least 4 dark chocalates
perl
  • perl
that produces (16,4) (15, 5 ) (14,6) (13, 7) (12,8)
perl
  • perl
the situation is constrained by two facts you cannot have more than 16 non-dark chocolates you cannot have more than 8 dark chocolates there must be at least 4 dark chocalates so the possibilities are (# non-dark, # dark) (16,4) (15, 5 ) (14,6) (13, 7) (12,8) there is 4/5 favorable
perl
  • perl
can i get a yes?
anonymous
  • anonymous
sure, i guess
perl
  • perl
lol
anonymous
  • anonymous
the explanation seems reasonable enough
perl
  • perl
i think its more complicated, let X = number of dark ones , Y = number of non-dark chocolate how many ways can you choose 16 non dark and 4 dark? how many ways can you choose 15 non-dark and 5 dark?
perl
  • perl
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 )
anonymous
  • anonymous
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
perl
  • perl
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) + ...
anonymous
  • anonymous
btw, what level of maths is this?
perl
  • perl
not sure
anonymous
  • anonymous
i meant like are you still in school or university?
perl
  • perl
P(Y=16 & X=4) = P(Y=16) * P (X=4 | Y = 16) =
perl
  • perl
im in university
anonymous
  • anonymous
oh, then you might not want to listen to me, i'm in yr 11 doing yr 12 maths so yea...
perl
  • perl
i find this problem confusing
anonymous
  • anonymous
whoops, should have told you that at the start!
perl
  • perl
im trying to force these into mutually exclusive possibilities
perl
  • perl
you cannot have more than 16 non-dark chocolates you cannot have more than 8 dark chocolates there must be at least 4 dark chocalates so the possibilities are (# non-dark, # dark) (16,4) (15, 5 ) (14,6) (13, 7) (12,8)
perl
  • perl
but are these equally likely?
perl
  • perl
if these possibilities are equally likely to occur (and they are mutually exclusive), then the probability is 4/5
anonymous
  • anonymous
i'd assume so, unless some were heavier or bigger than the others (which isn't so)
perl
  • perl
im use to working on problems like 5/12*4/11 , those sorts of problems
anonymous
  • anonymous
me too, I guess you could do that, but you'd be there forever!
perl
  • perl
what was your approach?
anonymous
  • anonymous
I found the probability of getting all 8 chocolates, kinda took a while and subtracted it from 1
perl
  • perl
can you show me your work
perl
  • perl
i need your help :)
perl
  • perl
|dw:1366635152002:dw|
anonymous
  • anonymous
hmm..i guess you could use combinatorics
anonymous
  • anonymous
does order matter?
perl
  • perl
because i want Probability ( 4 dark& 16 non-dark) exactly
perl
  • perl
\[\frac{ 8C4*16C16 }{24C20 }+\frac{ 8C5*16C15 }{24C20}+\frac{ 8C6*16C14}{24C20}+\frac{ 8C7*16C13 }{24C20 }\]
anonymous
  • anonymous
what do you get from that?
perl
  • perl
the first is the case 4 dark & 16 non-dark then 5 dark& 15 nondark
perl
  • perl
the order of eating them does not count, but we should label in our mind the chocolates. if you want, label the dark chocolates 1-8, and label the non-dark chocolates 9-24
perl
  • perl
so now we have the problem dark chocolate = { 1,2,3,4,5,6,7,8} , non-dark { 9,10,11,12,... 23,24} , the numbers are going to be labels for the chocolates, ok ?
perl
  • perl
i could have used letters,
anonymous
  • anonymous
sure
perl
  • perl
so for instance, suppose we want four dark chocolates, and 16 non-dark 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 non-dark from 16 non dark. there is only one way. so im using multiplication rule (and probability)
perl
  • perl
the denominator is the total number of ways to choose 20 chocolates , and there are 24 choose 20 ways to choose 20 chocolates
perl
  • perl
anyways, show me your work, and we can compare answers . sorry if my explanation is confusing, im not exactly sure of the correct jargon
perl
  • perl
so I got a probability of 629/759 = .8287
anonymous
  • anonymous
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
anonymous
  • anonymous
it's close to 4/5 at least
perl
  • perl
thats right, and yes its close to 4/5, but 4/5 would give you a wrong answer. i know how teachers are, lol
anonymous
  • anonymous
perl your answer is right thx
perl
  • perl
and you could solve it using the complement approach .
perl
  • perl
kabric, how do you know?
anonymous
  • anonymous
yea, basically what i was thinking
perl
  • perl
chelsea, how did you get nCr using pretty print?
anonymous
  • anonymous
i just typed it?
perl
  • perl
\[1-\frac{ \left(\begin{matrix}8 \\ 8\end{matrix}\right)*\left(\begin{matrix}16 \\ 12\end{matrix}\right) }{ \left(\begin{matrix}24 \\ 20\end{matrix}\right) }\]
perl
  • perl
that is 8C8 , 16 C 12
perl
  • perl
do you how it comes out the same ?
anonymous
  • anonymous
what do you mean?
anonymous
  • anonymous
you sort of lost me there
perl
  • perl
oh ok sorry
perl
  • perl
|dw:1366637463112:dw|
anonymous
  • anonymous
yes...
perl
  • perl
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
perl
  • perl
we cant have 9 dark ones, since there arent 9 dark ones
anonymous
  • anonymous
sure
perl
  • perl
ok lets not pursue this further, i have a new question answering, follow me
anonymous
  • anonymous
Correct answer => http://629a3486.linkbucks.com

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