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dont really get it

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

oh woops

i thought it said 'any'

do you get the question now?

yes

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

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

notice
3 dark 17 non-dark is impossible

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

can i get a yes?

sure, i guess

lol

the explanation seems reasonable enough

btw, what level of maths is this?

not sure

i meant like are you still in school or university?

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

im in university

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

i find this problem confusing

whoops, should have told you that at the start!

im trying to force these into mutually exclusive possibilities

but are these equally likely?

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

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

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

what was your approach?

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

can you show me your work

i need your help :)

|dw:1366635152002:dw|

hmm..i guess you could use combinatorics

does order matter?

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

what do you get from that?

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

i could have used letters,

sure

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

it's close to 4/5 at least

perl your answer is right thx

and you could solve it using the complement approach .

kabric, how do you know?

yea, basically what i was thinking

chelsea, how did you get nCr using pretty print?

i just typed it?

that is 8C8 , 16 C 12

do you how it comes out the same ?

what do you mean?

you sort of lost me there

oh ok sorry

|dw:1366637463112:dw|

yes...

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

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

sure

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

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