Q. What is the probability of getting a sum more than 7 when 2 dice are thrown?

• n(S)=6*6=36
  E=getting a sum more than 7 =>
  (2,6),(3,5),(4,4),(5,3),(6,2)
  (3,6),(4,5),(5,4),(6,3)
  (4,6),(5,5),(6,4)
  (5,6),(6,5)
  (6,6)
  n(E)=5+4+3+2+1=15
  
  p=n(E)/n(S)= 15/36 = 5/12
• 10 years agoHelpfull: Yes(21) No(0)
• total possible case of sum > 7 are (2,6) (3,5) (3,6) (4,4) (4,5) (4,6) (5,3) (5,4)
  (5,5) (5,6) (6,2) (6,3) (6,4) (6,5) (6,6) i.e., 15 possibal values and over all pairs are 6*6 =36 in number
  hence probability is 15/36 = 5/12
• 10 years agoHelpfull: Yes(5) No(0)
• 5+4++3+2+1=15/36
• 10 years agoHelpfull: Yes(2) No(0)
• it is more than 7 only not equal to 7
  so (2,6),(3,5),(3,6),(4,4),(4,5),(4,6),(5,3),(5,4),(5,5),(5,6),(6,2),(6,3),(6,4),(6,5),(6,6) total is=15=n(e)
  n(s)=36
  n(e)/n(s)=15/36=5/12
• 10 years agoHelpfull: Yes(1) No(0)
• 15
  =(2,6),(3,5),(3,6),(4,4)(4,5),(4,6),(5,3)(5,4)(5,5)(5,6)(6,2)(6,3)(6,4)(6,5)(6,6)
• 10 years agoHelpfull: Yes(1) No(0)
• (2,6)(6,2)
  (3,5)(5,3)(3,6)(6,3)
  (4,4)(4,5)(5,4)(4,6)(6,4)
  (5,5)
  (6,5)(5,6)
  (6,6)
  (5,6)(6,5)
  15/36=5/12
• 10 years agoHelpfull: Yes(1) No(0)
• ans:7/12
  (6+5+.....+1)/36
• 10 years agoHelpfull: Yes(0) No(6)
• probability of getting less and equal to seven is=19/36
  so
  1-19/36=17/36
  and 17/36
• 10 years agoHelpfull: Yes(0) No(0)