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

- schrodinger

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- anonymous

dont really get it

- UsukiDoll

choose milk chocolates choose white chocolates and just 4 dark chocolates XD

- 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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## More answers

- perl

oh woops

- perl

i thought it said 'any'

- anonymous

do you get the question now?

- perl

yes

- UsukiDoll

I'm thinking
milk white milk white milk white milk white milk white

- 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

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

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

since they must add up to 20 , these are the only possibilites

- perl

notice
3 dark 17 non-dark is impossible

- 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

that produces (16,4) (15, 5 ) (14,6) (13, 7) (12,8)

- 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

can i get a yes?

- anonymous

sure, i guess

- perl

lol

- anonymous

the explanation seems reasonable enough

- 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

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

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

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

btw, what level of maths is this?

- perl

not sure

- anonymous

i meant like are you still in school or university?

- perl

P(Y=16 & X=4) = P(Y=16) * P (X=4 | Y = 16) =

- perl

im in university

- anonymous

oh, then you might not want to listen to me, i'm in yr 11 doing yr 12 maths so yea...

- perl

i find this problem confusing

- anonymous

whoops, should have told you that at the start!

- perl

im trying to force these into mutually exclusive possibilities

- 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

but are these equally likely?

- perl

if these possibilities are equally likely to occur (and they are mutually exclusive), then the probability is 4/5

- anonymous

i'd assume so, unless some were heavier or bigger than the others (which isn't so)

- perl

im use to working on problems like 5/12*4/11 , those sorts of problems

- anonymous

me too, I guess you could do that, but you'd be there forever!

- perl

what was your approach?

- anonymous

I found the probability of getting all 8 chocolates, kinda took a while and subtracted it from 1

- perl

can you show me your work

- perl

i need your help :)

- perl

|dw:1366635152002:dw|

- anonymous

hmm..i guess you could use combinatorics

- anonymous

does order matter?

- perl

because i want Probability ( 4 dark& 16 non-dark) exactly

- perl

\[\frac{ 8C4*16C16 }{24C20 }+\frac{ 8C5*16C15 }{24C20}+\frac{ 8C6*16C14}{24C20}+\frac{ 8C7*16C13 }{24C20 }\]

- anonymous

what do you get from that?

- perl

the first is the case 4 dark & 16 non-dark
then 5 dark& 15 nondark

- 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

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

i could have used letters,

- anonymous

sure

- 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

the denominator is the total number of ways to choose 20 chocolates , and there are 24 choose 20 ways to choose 20 chocolates

- 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

so I got a probability of 629/759 = .8287

- 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

it's close to 4/5 at least

- 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

perl your answer is right thx

- perl

and you could solve it using the complement approach .

- perl

kabric, how do you know?

- anonymous

yea, basically what i was thinking

- perl

chelsea, how did you get nCr using pretty print?

- anonymous

i just typed it?

- 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

that is 8C8 , 16 C 12

- perl

do you how it comes out the same ?

- anonymous

what do you mean?

- anonymous

you sort of lost me there

- perl

oh ok sorry

- perl

|dw:1366637463112:dw|

- anonymous

yes...

- 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

we cant have 9 dark ones, since there arent 9 dark ones

- anonymous

sure

- perl

ok lets not pursue this further, i have a new question answering, follow me

- anonymous

Correct answer => http://629a3486.linkbucks.com

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