Three dice are tolled what is the probability of getting the sum as 13?

• ANS: n(S)=6^3=216
  n(E)={(1,6,6),(6,6,1),(6,1,6),(2,6,5),(2,5,6),(6,2,5),(3,6,5),(6,3,5),(6,5,3),(3,5,6),(5,6,3),(5,3,6),(4,6,3),(6,4,3),(6,3,4),(4,3,6),(3,6,4),(3,4,6),(5,6,2),(6,5,2),(5,2,6)} =21
  Probability of getting the sum as 13 is n(E)/n(S)=21/216.
  
• 9 years agoHelpfull: Yes(6) No(1)
• 21/216
  ((20-n)C2 )/216 where n=13
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• 21/216.
  Make Group of 13 Which is 21.
  
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• 1/18
  is the ans
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• 9/(6*6*6)
  =1/24
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• p(E)= n(E)/n(S)
  17/216
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• p(e)=n(e)/n(s)
  p(getting 13 as sum)=18/216
  ans=18/216
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• 661- 6combinations
  652-6 combinations
  643-6combinations
  553-6combinations
  544-6combinations
  
  total 36 combination i.e n(E)=36,n(s)=6^6^6
  ===> answer 36/216
• 9 years agoHelpfull: Yes(0) No(1)
• 13/216
  n(e)=
  (1,6,6),(6,6,1),(6,1,6),(2,6,5),(2,5,6),(6,2,5)(4,6,3),(6,4,3),(6,3,4),(4,3,6),(3,6,4),(3,4,6),(5,6,2),(6,5,2),(5,2,6)
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• 21/216 is the answer
  becoz 6+5+4+3+2+1=21
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• #Gowthami,u r wrong,why u calculate sum=14. here sum=13 is in questions.
  15/(6*6*6)=5/108
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