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fav no. of outcomes =4096, I ran a c++ program for that, in case anyone would like to view it ?
am not too sure with the sample space.
i am not sure but one version of the sample space is just all possible \(6^6\) outcomes when you roll six dice
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.
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
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
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
oh geez, yes i do! seems very very complicated
i would have no idea how to do this
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
evidently you have to exclude any 3 and any 6, but after that i cannot imagine how you proceed
yeah what you said is what i think it is
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
man good luck!!
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
Didn't it help?
I never really followed :/
Should I post the whole solution?
Actually I found a solution on yahooanswers, didn't follow that either :P
wait, I shall post the link ^_^
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
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
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
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
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.
Please give me some time to grasp.
Am so sorry could not understand, my head aches! :( I'll give it a fresh start later :| so sorry :/
As u wish
Thanks a ton sire! I was revising this and everything is crystal clear now! ^_^ Can't thank you enough..