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 one year ago
I probably have the correct number of favorable outcomes for this, but can anyone tell me what should be the sample space ?
 one year 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
 one year 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
 one year ago
Best ResponseYou've already chosen the best response.0am not too sure with the sample space.

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1i am not sure but one version of the sample space is just all possible \(6^6\) outcomes when you roll six dice

shubhamsrg
 one year 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.

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1well not really, the sample space is usually defined to be "the set of all possible outcomes" but not necessarily the denominator of your probability

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1the 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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1but 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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1oh geez, yes i do! seems very very complicated

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1i would have no idea how to do this

shubhamsrg
 one year 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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1evidently you have to exclude any 3 and any 6, but after that i cannot imagine how you proceed

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1yeah what you said is what i think it is

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1so 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

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.2Hint: 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
 one year ago
Best ResponseYou've already chosen the best response.0I never really followed :/

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.2Should I post the whole solution?

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

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

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

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.2when 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

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.2Now, 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

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.2So, 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

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.2Since, 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

sauravshakya
 one year ago
Best ResponseYou've already chosen the best response.2Now, 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
 one year ago
Best ResponseYou've already chosen the best response.0Please give me some time to grasp.

shubhamsrg
 one year 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
 one year 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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