when 3 dice are rolled find the probability of getting a sum of 13

• =21/(6^3)
  =7/72
• 12 years agoHelpfull: Yes(29) No(7)
• probability of getting the sum os 13 =
  21/216.
• 12 years agoHelpfull: Yes(28) No(6)
• 6,6,1= 3!/2!=3
  6,4,3=3!=6
  6,5,2=3!=6
  5,5,3=3!/2!=3
  5,4,4=3!/2!=3
  total cases= 3+6+6+3+3=21
  probability= 21/216
• 9 years agoHelpfull: Yes(18) No(1)
• trick to solve problems related to 3 dices:
  (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
• 12 years agoHelpfull: Yes(7) No(8)
• possibilities are
  166 616 661
  265 625 652
  256 526 562
  355 535 553
  463 643 634
  436 346 364
  544 454 445
  643 364 436
  634 463 346
  
  so 27 possiblities and total ways are 6^3=216
  so answer=27/216=1/8
  
• 12 years agoHelpfull: Yes(4) No(22)
• msantosh is crrct
• 11 years agoHelpfull: Yes(0) No(1)
• cases possible:- 6+6+1(3 cases)
  6+5+2(6cases)
  6+4+3(6cases)
  5+5+3(3cases)
  5+4+4(3cases)
  
  total 21 cases possible. hence p(sum=13)=21/216=7/72
• 9 years agoHelpfull: Yes(0) No(1)
• 13/16.13/(13 + 3)=16
• 9 years agoHelpfull: Yes(0) No(0)