## shubhamsrg 2 years ago I probably have the correct number of favorable outcomes for this, but can anyone tell me what should be the sample space ?

1. shubhamsrg
2. shubhamsrg

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

3. shubhamsrg

am not too sure with the sample space.

4. satellite73

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

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. satellite73

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. satellite73

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. satellite73

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. satellite73

oh geez, yes i do! seems very very complicated

10. satellite73

i would have no idea how to do this

11. shubhamsrg

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. satellite73

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

13. satellite73

yeah what you said is what i think it is

14. satellite73

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

yep.

16. satellite73

man good luck!!

17. shubhamsrg

:D

18. sauravshakya

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

Didn't it help?

20. shubhamsrg

I never really followed :/

21. sauravshakya

Should I post the whole solution?

22. shubhamsrg

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

23. ParthKohli

Which one?

24. shubhamsrg

wait, I shall post the link ^_^

25. shubhamsrg
26. ParthKohli

Seems familiar.

27. shubhamsrg

hmm..

28. shubhamsrg

@sauravshakya ?

29. sauravshakya

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

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

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

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

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

Please give me some time to grasp.

35. sauravshakya

okay

36. shubhamsrg

Am so sorry could not understand, my head aches! :( I'll give it a fresh start later :| so sorry :/

37. sauravshakya

As u wish

38. shubhamsrg

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