3 dice are rolled. Then what is the probability u will get the sum of number as 10.

• (6,3,1)=>6 ways
  (6,2,2)=>3ways
  (5,4,1)=>6ways
  (5,2,3)=>6ways
  (4,2,4)=>3ways
  (4,3,3)=>3ways
  no of ways to sum=10=> n(E) 27ways
  n(s)=6^3=216
  p=n(E)/n(S)=27/216
  p=1/8
• 11 years agoHelpfull: Yes(33) No(0)
• Ans: 1/8
  (1,3,6) => 3!=6
  (1,4,5) => 3!=6
  (2,2,6) => 3!/2=3
  (2,3,5) => 3!=6
  (2,4,4) => 3!/2=3
  (3,3,4) => 3!/2=3
  Total = 3+3+3+6+6+6 = 27
  Hence, 27/216 = 1/8...
  
• 11 years agoHelpfull: Yes(21) No(0)
• no. of chances to get sum as 10 are
  (6,3,1)=3! ways
  (6,2,2)=3!/2
  (5,4,1)=3!
  (5,3,2)=3!
  (4,3,3)=3!/2
  (4,2,4)=3!/2
  so no.of chances= 6+3+6+6+3+3=27
  hence probability to get sum as 10 is= 27/216
• 11 years agoHelpfull: Yes(6) No(1)
• 18/216
  =1/12
• 11 years agoHelpfull: Yes(2) No(15)
• the answer is 1/8 by the same logic already explained.
  when 3 die r thrown with any 2 having the same no. face up, then the no. of ARRANGEMENTS possible is 3!/2!.
  its a perm com formula.
  no. of arrangements possible with n items where k are of the same type= n!/k!
• 11 years agoHelpfull: Yes(1) No(0)
• Hey guys i have doubt who posted 1/8 as answer.
  how can take repeated case, consider below
  getting (2,2) on throwing 2 dice probability is 2/36 not 1/36.
  If I am wrong please clarify me.
• 11 years agoHelpfull: Yes(0) No(5)
• 27/216...by considering al the possibilities like 3,2,5..6,2,2..4,2,4..6,3,1.. etc
• 11 years agoHelpfull: Yes(0) No(1)
• 5*3!/216=30/216
• 11 years agoHelpfull: Yes(0) No(2)