If four dice are thrown and the values are noted. What is the probability of the sum being 20?

• (6,6,6,2)=4!/3!=4
  (6,6,5,3)=4!/2!=12
  (6,6,4,4)=4!/(2!*2!)=6
  (6,5,5,4)=4!/2!=12
  (5,5,5,5)=1
  therefore ans will be 35/1296
• 11 years agoHelpfull: Yes(55) No(0)
• there will be one more 5,5,5,5.. so 34+1 =35
• 11 years agoHelpfull: Yes(7) No(3)
• possible combi
  (6,6,6,2)=(4!/3!)=4
  (6,6,4,4)=(4!/(2!*2!))=6.
  (6,6,5,3)=(4!/2!)=12.
  (6,5,5,4)=(4!/2!)=12.
  
  prob=(4+6+12+12)/(6^4) = 34/1296
• 11 years agoHelpfull: Yes(6) No(13)
• a+b+c+d = 20,
  
  there are total 35 ways of values of a,b,c,d.
  therefore prob=35/36*36. Ans
• 11 years agoHelpfull: Yes(1) No(3)
• (6,6,6,2)=4!/3!=4
  (6,6,5,3)=4!/2!=12
  (6,6,4,4)=4!/(2!*2!)=6
  (6,5,5,4)=4!/2!=12
  (5,5,5,5)=1
  therefore ans will be 35/1296
• 11 years agoHelpfull: Yes(1) No(3)