2014 tcs ques- In a single throw with two dice, find the probability that their sum is a multiple of 3 or 4

• Possible sums: 3,4,6,8,9,12
  3=(1,2),(2,1)
  4=(1,3),(2,2),(3,1)
  6=(1,5),(2,4),(3,3),(4,2),(5,1)
  8=(2,6),(3,5),(4,4),(5,3),(6,2)
  9=(3,6),(4,5),(5,4),(6,3)
  12=(6,6)
  
  So possible outcomes = 20
  So probability= 20/6^2 = 5/9
• 9 years agoHelpfull: Yes(42) No(0)
• possible sums:{(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)}
  p(n)=20
  p(s)=36
  probability(sum is a multiple of 3 or 4)=20/36=5/9
• 9 years agoHelpfull: Yes(6) No(0)
• Possible sums: 3,4,6,8,9,12
  3=(1,2),(2,1)
  4=(1,3),(2,2),(3,1)
  6=(1,5),(2,4),(3,3),(4,2),(5,1)
  8=(2,6),(3,5),(4,4),(5,3),(6,2)
  9=(3,6),(4,5),(5,4),(6,3)
  12=(6,6)
  
  So possible outcomes = 20
  So probability= 20/6^2 = 5/9
• 9 years agoHelpfull: Yes(4) No(0)
• 20/36=5/9 20/36 is crct i think.
• 9 years agoHelpfull: Yes(2) No(0)
• p(3)=11/36
  p(4)=8/36
  p(3^4)=1/36
  required probability=p(3)+p(4)-p(3^4)
  =1/2
• 9 years agoHelpfull: Yes(2) No(2)
• Posible ways - 2+3+5+5+4+1 = 20
  
  p(20/6*6)= (5/9)
• 9 years agoHelpfull: Yes(2) No(0)
• two dice are thrown = 36
  sum is a multiple of 3 or 4 = {(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)} = 19
  
  probability = 19/36
• 9 years agoHelpfull: Yes(0) No(5)
• p(3)=5/36
  p(4)=5/36
  p(3,4)=1/36
  ans=5/36+5/36-1/36
  =1/4
• 9 years agoHelpfull: Yes(0) No(2)
• 21/36 IS ANSWER bcuz 21 posssible sums can be obtained and if two dice are thrown together then possible events will be 36 thus answer is 21/36
• 9 years agoHelpfull: Yes(0) No(3)
• 20/36 is the right answer
  
• 9 years agoHelpfull: Yes(0) No(2)
• total outcome=36
  possible outcome=20
  so probability is 20/36=5/9
• 9 years agoHelpfull: Yes(0) No(0)
• Ans is 5/9
• 9 years agoHelpfull: Yes(0) No(0)
• Possible Sum: 3,4,6,8,9,12
  Possible Outcomes: 20
  Total outcome: 36
  Probability: 20/36=5/9
• 9 years agoHelpfull: Yes(0) No(0)
• probablity for multiple of 3=12/36
  probablity for multiple of 4=9/36
  total=12/36+9/36
  ans=7/12
• 9 years agoHelpfull: Yes(0) No(0)
• 5/36. Probability of getting 3 is 2/36, Probability of getting 4 is 3/36. Here either 3 or 4 so add them we get 5/36
• 9 years agoHelpfull: Yes(0) No(0)
• Possible sums-3,4,6,8,9,12
  3--(1,2)
  4--(1,3) , (2,2)
  6--(1,5),(2,4),(3,3)
  8--(2,6),(3,5),(4,4)
  9--(3,6),(4,5)
  12--(6,6)
  So possible outcomes--12
  Total outcomes--6*6=36
  So probability=12/36=1/3
• 9 years agoHelpfull: Yes(0) No(2)
• Throwing two dice so total outcomes is 6*6=36
  No of possible outcomes is
  (1,2,3,4,5,6,)
  (1,2,3,4,5,6)
  no of outcomes
  (1,2)(1,3)(1,5)==3
  (2,1)(2,2)(2,4)(2,6)==4
  (3,1)(3,3)(3,5)(3,6)==4
  (4,2)(4,4)(4,5)==3
  (5,1)(5,3)(5,4)==3
  (6,2)(6,3)(6,6)==3
  totol 20
  so
  20/36 = 5/9
  ans is 5/9
• 9 years agoHelpfull: Yes(0) No(0)