Question

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

Answer 1

http://gyazo.com/7df2a488ddba6e26108b9830d264fa5b

Answer 2

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

Answer 3

am not too sure with the sample space.

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

oh geez, yes i do! seems very very complicated

Answer 10

i would have no idea how to do this

Answer 11

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

Answer 12

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

Answer 13

yeah what you said is what i think it is

Answer 14

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

Answer 15

yep.

Answer 16

man good luck!!

Answer 17

:D

Answer 18

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

Answer 19

Didn't it help?

Answer 20

I never really followed :/

Answer 21

Should I post the whole solution?

Answer 22

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

Answer 23

Which one?

Answer 24

wait, I shall post the link ^_^

Answer 25

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

Answer 26

Seems familiar.

Answer 27

hmm..

Answer 28

@sauravshakya ?

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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.

Answer 34

Please give me some time to grasp.

Answer 35

okay

Answer 36

Am so sorry could not understand, my head aches! :(

I'll give it a fresh start later :|
so sorry :/

Answer 37

As u wish

Answer 38

Thanks a ton sire! I was revising this and everything is crystal clear now! ^_^

Can't thank you enough..