If two dice are rolled to obtain a sum or 5 or 7 what is the probability of getting 5 before 7?

• Total no. of outcomes to obtain 5 or 7 is 10 on rolled of two dice. getting sum 5 before sum 7 is 4.so,
  prob=4/10 i.e 2/5
• 10 years agoHelpfull: Yes(9) No(2)
• 2/5
  4/36 / 4/36+6/36
• 11 years agoHelpfull: Yes(6) No(2)
• At each throw we have:
  
  Probability of 5 = 4/36 = 1/9 Probability of 7 = 6/36 = 1/6
  
  Probability of no result = 1 - (1/9 + 1/6)
  
  = 13/18
  :
  now probability for getting 5 or 7 as sum wd gtting 5 before is
  
  p = 1/9 + (13/18)(1/9) + (13/18)^2(1/9) + (13/18)^3(1/9) + ...
  
  = (1/9)[1 + 13/18 + (13/18)^2 + (13/18)^3 + ...... ]
  
  This is a GP with common ratio 13/18
  
  = (1/9)/[1 -13/18]
  
  = (1/9)/(5/18)
  
  = 2/5
  
  A
• 10 years agoHelpfull: Yes(6) No(0)
• probability of getting a before b is =p(a)/(p(a)+p(b))
  here p(a)=4/36
  p(b)=p(7)=6/36
  ans=2/5
• 9 years agoHelpfull: Yes(2) No(0)
• probability of getting a sum of 5 or 7 is 10/36
  and probability of getting 5 before 7 is 4/36 = 1/9
• 10 years agoHelpfull: Yes(1) No(2)