Raj tossed three dice and their results are noted down. Find the total number of ways in which Raj can get the sum as 13. Also find the probability of get the sum as 10.

• (6,6,1)=3!/2!=3
  (6,5,2)=3!=6
  (6,4,3)=3!=6
  (5,5,3)=3!/2=3
  (5,4,4)=3!/2!=3
  total=21
  P(sum 13)=21/216
• 11 years agoHelpfull: Yes(42) No(2)
• Always remember when 3 dice are rolled the number of ways of getting n ( where n is the sum of faces on dice)
  = (n−1)C2 where n = 3 to 8
  = 25 where n = 9, 12
  = 27 where n = 10, 11
  = (20−n)C2 where n = 13 to 18
  The required probability for sum 10 is = 27/6^3 = 27/216
  For sum 13 (20-13)C2=7C2=21[ where n=13]
  The required probability for sum 13 is = 21/6^3 = 21/216= 7/72
  
• 11 years agoHelpfull: Yes(33) No(4)
• number of ways in which 13 can appear on three dices
  (6,6,1) = 3! / 2! = 3
  (6,5,2) = 3! = 6
  (6,4,3) = 3! = 6
  (5,5,3) = 3! / 2! = 3
  (5,4,4) = 3! / 2! = 3
  (6,3,4) = 3! = 6
  So the total number of ways in which 13 can appear will be 3+6+6+3+3+6 = 27
• 11 years agoHelpfull: Yes(7) No(23)
• no of ways of 13=21;
  p(10)=27/216;
• 11 years agoHelpfull: Yes(2) No(2)
• i think it shud be 21*3/216
  
• 11 years agoHelpfull: Yes(0) No(0)
• i think it is 18
  
• 10 years agoHelpfull: Yes(0) No(0)