two dice are thrown. find the probability of getting a multiple of 3 of 4 as the sum

• 20/36=5/9
  (1,2),(1,3),(1,5),(2,1)(2,2)(2,4)(2,6)(3,1)(3,3)(3,5)(3,6)(4,2)(4,4)(4,5)(5,1)(5,3)(5,4)(6,2)(6,3)(6,6) as its some is multiple of 3 or 4
• 8 years agoHelpfull: Yes(39) No(0)
• Their sum can be 3,4,6,8,9,12
  For two dice, any number from 2 to 7 can be get in (n-1) ways and any number from 8 to 12 can be get in (13 - n) ways.
  Then possible ways are 2 + 3 + 5 + 5 + 4 + 1 = 20 possible cases.
  So probability is (20/36)=(5/9)
• 8 years agoHelpfull: Yes(4) No(1)
• (1,2)(1,3)(1,5)
  (2,1)(2,2)(2,4)(2,8)
  (3,1)(3,3)(3,5)(3,6)
  (4,2)(4,4)(4,5)
  (5,1)(5,3)(5,4)
  (6,2)(6,3)(6,6)
  so total outcomes will be 20
  probablity=20/36=5/9
• 8 years agoHelpfull: Yes(3) No(0)
• 23/36
  Solution
• 8 years agoHelpfull: Yes(1) No(9)
• Probability – Sample space for two dice (outcomes):
  (1,1)(1,2)(1,3)(1,4)(1,5)(1,6)
  (2,1)(2,2)(2,3)(2,4)(2,5)(2,6)
  (3,1)(3,2)(3,3)(3,4)(3,5)(3,6)
  (4,1)(4,2)(4,3)(4,4)(4,5)(4,6)
  (5,1)(5,2)(5,3)(5,4)(5,5)(5,6)
  (6,1)(6,2)(6,3)(6,4)(6,5)(6,6)
  
  multiple of 3 of 4 as the sum =
  
• 8 years agoHelpfull: Yes(1) No(6)
• 1/2
  ....18/36
  
• 8 years agoHelpfull: Yes(0) No(11)
• 18/36=1/2 multiple of 3 or 4 is 18
  and total no of throws is 36
• 8 years agoHelpfull: Yes(0) No(6)
• 5= (1,2),(2,1),(1,3),(3,1),(2,2)
  5/36
• 8 years agoHelpfull: Yes(0) No(3)
• thank u abhineet kumar
  
• 8 years agoHelpfull: Yes(0) No(0)
• 2/12 + 3/12 = 5/12
  Is it correct?
• 8 years agoHelpfull: Yes(0) No(2)
• As I think
  That in Question two die should be multiple of 3 and which should 4 as a addition
  So
  Number will be : ( 2,4) , (5,4)
  Then 2/36
  Will gives 1/18
• 5 years agoHelpfull: Yes(0) No(0)
• 20/36=5/9
  (1,2),(1,3),(1,5),(2,1)(2,2)(2,4)(2,6)(3,1)(3,3)(3,5)(3,6)(4,2)(4,4)(4,5)(5,1)(5,3)(5,4)(6,2)(6,3)(6,6) as its some is multiple of 3 or 4
• 1 year agoHelpfull: Yes(0) No(0)