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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 ?

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  1. shubhamsrg
    • 3 years ago
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    http://gyazo.com/7df2a488ddba6e26108b9830d264fa5b

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

  3. shubhamsrg
    • 3 years ago
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    am not too sure with the sample space.

  4. anonymous
    • 3 years ago
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    i am not sure but one version of the sample space is just all possible \(6^6\) outcomes when you roll six dice

  5. shubhamsrg
    • 3 years ago
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    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.

  6. anonymous
    • 3 years ago
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    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

  7. anonymous
    • 3 years ago
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    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

  8. anonymous
    • 3 years ago
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    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

  9. anonymous
    • 3 years ago
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    oh geez, yes i do! seems very very complicated

  10. anonymous
    • 3 years ago
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    i would have no idea how to do this

  11. shubhamsrg
    • 3 years ago
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    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

  12. anonymous
    • 3 years ago
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    evidently you have to exclude any 3 and any 6, but after that i cannot imagine how you proceed

  13. anonymous
    • 3 years ago
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    yeah what you said is what i think it is

  14. anonymous
    • 3 years ago
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    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

  15. shubhamsrg
    • 3 years ago
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    yep.

  16. anonymous
    • 3 years ago
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    man good luck!!

  17. shubhamsrg
    • 3 years ago
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    :D

  18. sauravshakya
    • 3 years ago
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    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

  19. sauravshakya
    • 3 years ago
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    Didn't it help?

  20. shubhamsrg
    • 3 years ago
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    I never really followed :/

  21. sauravshakya
    • 3 years ago
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    Should I post the whole solution?

  22. shubhamsrg
    • 3 years ago
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    Actually I found a solution on yahooanswers, didn't follow that either :P

  23. ParthKohli
    • 3 years ago
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    Which one?

  24. shubhamsrg
    • 3 years ago
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    wait, I shall post the link ^_^

  25. shubhamsrg
    • 3 years ago
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    http://answers.yahoo.com/question/index?qid=20130312081947AAUNJ7K

  26. ParthKohli
    • 3 years ago
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    Seems familiar.

  27. shubhamsrg
    • 3 years ago
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    hmm..

  28. shubhamsrg
    • 3 years ago
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    @sauravshakya ?

  29. sauravshakya
    • 3 years ago
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    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

  30. sauravshakya
    • 3 years ago
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    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

  31. sauravshakya
    • 3 years ago
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    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

  32. sauravshakya
    • 3 years ago
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    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

  33. sauravshakya
    • 3 years ago
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    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.

  34. shubhamsrg
    • 3 years ago
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    Please give me some time to grasp.

  35. sauravshakya
    • 3 years ago
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    okay

  36. shubhamsrg
    • 3 years ago
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    Am so sorry could not understand, my head aches! :( I'll give it a fresh start later :| so sorry :/

  37. sauravshakya
    • 3 years ago
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    As u wish

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

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