Three dice are rolled. What is the probability that you will get the sum of the numbers as 10?

• n(s)=6^3=216.
  no. of ways to get 10 as sum r
  (2,2,6) -> 3!/2!=3.
  (2,4,4) -> 3!/2!= 3
  (1,3,6) -> 3!=6
  (1,4,5) -> 3!=6
  (2,3,5) -> 3!=6
  (3,3,4) -> 3!/2!=3
  Total n(e)=27.
  Probability=n(e)/n(s)=27/216=1/8.
• 11 years agoHelpfull: Yes(28) No(1)
• 27/216
  Let E be the event to get sum as 10
  So E= {136,145,154,163,
  226,235,244,253,262
  316,325,334,343,352,361,
  415,424,433,442,451
  514,523,532,541
  613,622,631}
  N(E)= 27
  now n(s) = 216. ( here s sample space )
  so p(E)= 27/216=1/8
• 11 years agoHelpfull: Yes(2) No(0)
• total sample space=216
  total event=27
  so probability=27/216
• 11 years agoHelpfull: Yes(0) No(2)
• total sample space=216
  total event=27
  so probability=27/216
• 11 years agoHelpfull: Yes(0) No(2)
• basically its solving for equation x+y+z=10,
  where x,y,z can take any value of 1,2,3,4,5,6;
  
  solving this equation :
  
  think x+y+z =10 like this,
  number of ways we can distribute each 1 (10 one's ) by 2 sticks(sticks represent + sign)
  
  e.g. 1 1 1 | 1 1 1 1 | 1 1 1
  gap is filled by 2 sticks in this example making a possible permuation as 3+3+2;
  
  1 Gap 1 Gap 1 Gap 1 Gap 1 Gap 1 Gap 1 Gap 1 Gap 1 Gap 1
  
  so we can chose 2 gaps out of 9 gaps in 9C2 ways.
  
  but
  1+1+8
  1+8+1
  8+1+1
  1+2+7 (permuate it in 3!=6 ways)
  so 3 +9 of them must be subtracted out from 9C2.
  
  9C2-9=27 ways.
  
  probablity= 27/216.
• 4 years agoHelpfull: Yes(0) No(0)