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

This Question is Closed

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0fav no. of outcomes =4096, I ran a c++ program for that, in case anyone would like to view it ?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0am not too sure with the sample space.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i am not sure but one version of the sample space is just all possible \(6^6\) outcomes when you roll six dice

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0yep 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
 3 years ago
Best ResponseYou've already chosen the best response.0well 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
 3 years ago
Best ResponseYou've already chosen the best response.0the 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
 3 years ago
Best ResponseYou've already chosen the best response.0but 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
 3 years ago
Best ResponseYou've already chosen the best response.0oh geez, yes i do! seems very very complicated

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i would have no idea how to do this

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0hmm 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
 3 years ago
Best ResponseYou've already chosen the best response.0evidently you have to exclude any 3 and any 6, but after that i cannot imagine how you proceed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah what you said is what i think it is

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hint: 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

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0I never really followed :/

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Should I post the whole solution?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0Actually I found a solution on yahooanswers, didn't follow that either :P

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0wait, I shall post the link ^_^

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0http://answers.yahoo.com/question/index?qid=20130312081947AAUNJ7K

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0when 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
 3 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 3 years ago
Best ResponseYou've already chosen the best response.0So, 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
 3 years ago
Best ResponseYou've already chosen the best response.0Since, 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
 3 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 3 years ago
Best ResponseYou've already chosen the best response.0Please give me some time to grasp.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0Am so sorry could not understand, my head aches! :( I'll give it a fresh start later : so sorry :/

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0Thanks a ton sire! I was revising this and everything is crystal clear now! ^_^ Can't thank you enough..
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