I probably have the correct number of favorable outcomes for this, but can anyone tell me what should be the sample space ?

- shubhamsrg

- schrodinger

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- shubhamsrg

http://gyazo.com/7df2a488ddba6e26108b9830d264fa5b

- shubhamsrg

fav no. of outcomes =4096, I ran a c++ program for that, in case anyone would like to view it ?

- shubhamsrg

am not too sure with the sample space.

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## More answers

- anonymous

i am not sure but one version of the sample space is just all possible \(6^6\) outcomes when you roll six dice

- shubhamsrg

yep that was my first attempt, which leads us to 64/729 i.e. ans = 793 , but it tells me the ans is wrong.
Sample space should be something else, as we are allowed to arrange.

- anonymous

well not really, the sample space is usually defined to be "the set of all possible outcomes" but not necessarily the denominator of your probability

- anonymous

the fact that you cannot distinguish between (a,b,c,d,e,f) from (a,d,b,e,c,f) does not make them the same outcome

- anonymous

but should shut up because i have read your problem now 6 or 7 times and i do not understand it. it seems like the probability would depend on k

- anonymous

oh geez, yes i do! seems very very complicated

- anonymous

i would have no idea how to do this

- shubhamsrg

hmm
please tell me if I am understanding the problem correctly myself :
we throw 6 dice
1st no. should not be div by 3
sum of 1st 2 should also not be div by 3
sum of 1st 3..
.
.
6

- anonymous

evidently you have to exclude any 3 and any 6, but after that i cannot imagine how you proceed

- anonymous

yeah what you said is what i think it is

- anonymous

so for example the ordering on the left in the example does not work, but the ordering on the right does, so that one would be considered a success

- shubhamsrg

yep.

- anonymous

man good luck!!

- shubhamsrg

:D

- anonymous

Hint:
Let x be sum of outcomes of all 6 die then,
Look for case x Mod 3=1
And case x Mod 3 = 2 only.....
This will simplify the problem

- anonymous

Didn't it help?

- shubhamsrg

I never really followed :/

- anonymous

Should I post the whole solution?

- shubhamsrg

Actually I found a solution on yahooanswers, didn't follow that either :P

- ParthKohli

Which one?

- shubhamsrg

wait, I shall post the link ^_^

- shubhamsrg

http://answers.yahoo.com/question/index?qid=20130312081947AAUNJ7K

- ParthKohli

Seems familiar.

- shubhamsrg

hmm..

- shubhamsrg

@sauravshakya ?

- anonymous

when x Mod 3=1 then,
out of the six numbers, u can pick 3 numbers such that their sum is "a" where "a "mod 3 =0 and the rest three numbers sum is b where b mod 3=1

- anonymous

Now, Let p=1Mod3=4Mod3=1 , q=2Mod3=5Mod3=2 , r=3Mod3=6Mod6=0
So,
to pick three numbers such their sum is divisible by 3,
r,r,r
r,p,q
p,p,p
q,q,q
to pick three numbers such that their is b where b mod 3=1,
r,r,p
r,q,q
p,p,q

- anonymous

So, x mod 3=1 only when,
r,r,r,r,r,p
r,r,r,r,q,q
r,r,r,p,p,q
r,p,q,r,r,p
r,p,q,r,q,q
r,p,q,p,p,q
p,p,p,r,r,p
p,p,p,r,q,q
p,p,p,p,p,q
q,q,q,r,r,p
q,q,q,r,q,q
q,q,q,p,p,q

- anonymous

Since, r=0, p =1,q=2
r,r,r,r,r,p = 0,0,0,0,0,1
r,r,r,r,q,q=0,0,0,0,2,2
r,r,r,p,p,q=0,0,0,1,1,2
r,p,q,r,r,p=0,0,0,1,1,2 ->This is repeated
r,p,q,r,q,q=0,0,1,2,2,2
r,p,q,p,p,q=0,1,1,1,2,2
p,p,p,r,r,p=0,0,1,1,1,1
p,p,p,r,q,q=0,1,1,1,2,2->This is repeated
p,p,p,p,p,q=1,1,1,1,1,2
q,q,q,r,r,p=0,0,1,2,2,2 ->This is repeated
q,q,q,r,q,q=0,2,2,2,2,2
q,q,q,p,p,q=1,1,2,2,2,2

- anonymous

Now, out of these ,
r,r,r,r,r,p = 0,0,0,0,0,1
r,r,r,r,q,q=0,0,0,0,2,2
r,r,r,p,p,q=0,0,0,1,1,2
r,p,q,r,q,q=0,0,1,2,2,2
r,p,q,p,p,q=0,1,1,1,2,2
p,p,p,r,r,p=0,0,1,1,1,1
p,p,p,p,p,q=1,1,1,1,1,2
q,q,q,r,q,q=0,2,2,2,2,2
q,q,q,p,p,q=1,1,2,2,2,2
which can u arrange such that the sum of the first k numbers is not divisible by 3.

- shubhamsrg

Please give me some time to grasp.

- anonymous

okay

- shubhamsrg

Am so sorry could not understand, my head aches! :(
I'll give it a fresh start later :|
so sorry :/

- anonymous

As u wish

- shubhamsrg

Thanks a ton sire! I was revising this and everything is crystal clear now! ^_^
Can't thank you enough..

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