Two dice are thrown simultaneously. What is the probability that the sum of the numbers shown on the two dices will be a prime number?

• number of sample solution=36
  number of possible outcomes=(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2)(3,4)(4,1)(4,3)(5,2)(5,6)(6,1)(6,5)=15
  p=15/36
• 8 years agoHelpfull: Yes(25) No(2)
• number of outcomes = 15 (sum of prime numbers)
  possible outcomes = 36
  p(A) = 15/36 = 5/12
• 8 years agoHelpfull: Yes(6) No(0)
• two dice thrown means 6*6=36
  sum of two dice is prime number means 2,3,5,7,11......,
  no of events={(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(5,6),(6,1),(6,5)}=15
  probability=no of events/36=15/36=5/12
  
• 7 years agoHelpfull: Yes(3) No(0)
• sum is 2,3,5,7 and 11
  possible combinations are
  1,1
  1,2
  2,1
  1,4,
  2,3,
  3,2,
  4,1
  ,5,6,
  6,5,
  1,6,
  2,5,
  3,4,
  6,1,
  5,2,
  4,3
  hence the probability to get sum as prime on the two dice is 15/36
• 6 years agoHelpfull: Yes(3) No(1)
• total number of solutions =36
  no of possible outcomes = (1,1),(1,2),(1,4),(1,6),(2,3),(2,5),(3,4),(5,6) = 8
  remaining (2,1),(4,1),(6,1),(3,2),(5,2),(4,3),(6,5) are repeated
  probability = 8/36
• 8 years agoHelpfull: Yes(2) No(13)
• 7/18.As there r 36 total cases..And 14 favourable cases .
• 8 years agoHelpfull: Yes(2) No(4)
• Clearly, n(S) = (6 x 6) = 36.
  
  Let E = Event that the sum is a prime number.
  
  Then E = { (1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3),
  (5, 2), (5, 6), (6, 1), (6, 5) }
  n(E) = 15.
  
  P(E) = n(E) = 15 = 5 .
  n(S) 36 12
  
• 8 years agoHelpfull: Yes(1) No(0)
• sum of dice= 2 3 4 5 6 7 8 9 10 11 12
  no.of possible= 1 2 3 4 5 6 5 4 3 2 1
• 8 years agoHelpfull: Yes(1) No(0)
• when u throw two dice .there will be 15 prime numbers ,so 36/15 = 5/12
• 5 years agoHelpfull: Yes(1) No(0)
• The Possible Combinations are (1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2)(3,4)(4,1)(4,3)(5,2)(5,6)(6,1)(6,5)=15 and total outcome 6*6=36, So answer will be 15/36=5/12.
  
• 8 years agoHelpfull: Yes(0) No(1)
• total number of outcomes=6*6=36
  there are pairs which have prime sum=(1,1),(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2)(3,4)(4,1)(4,3)(5,2)(5,6)(6,1)(6,5)
  total number of above pairs=15
  so,p=15/36 = 5/12
• 8 years agoHelpfull: Yes(0) No(0)
• Select 6 men from 12 i.e 12C6 =926
  Now select 1 women from 3 i.e 3C1 =3
  then add 926 + 3 = 929,which is the answer
• 8 years agoHelpfull: Yes(0) No(2)
• there is probability of getting sum prime case 1=1+2,1+4
  case 2= 2+3,2+1
  case 3 =3+2
  case 4=4+1
  probability=6/12=1/2
• 8 years agoHelpfull: Yes(0) No(2)
• sum i.e prime=2 3 5 7 11
  total combination =2,3,4.................12
  required probablity=5/11
• 7 years agoHelpfull: Yes(0) No(2)
• Rolling 2 dice gives a total of 36 possible outcomes.
  Here is the sample space:
  
  
  Prob of getting a sum of atleast 6 = Prob of getting sum >=6
  
  Following table shows the possible cases(enclosed in orange box) for getting the required sum.
  
  
  Prob[Required Sum] =
  P[S=6]+P[S=7]+P[S=8]+P[S=9]+P[S=10]+P[S=11]+P[S=12]
  
  = 5/36+6/36+5/36+4/36+3/36+2/36+1/36
  
  = 26/36
  
  = 0.722
  
  Therefore, Required probability = 0.722
• 5 years agoHelpfull: Yes(0) No(3)