A pair of fair dices rolled together till a sum of either 5 or 7 is obtained what is the probability that 5 comes before 7?

• Event when sum is 5 or 7,E1=(1,4),(1,6),(2,3),(2,5),(3,2),(3,4),(4,1),(4,3),(5,2),(6,1)
  n(E1)=10
  Event when 5 comes before 7,E2=(1,4),(2,3),(3,2),(4,1)
  n(e2)=4
  So,P(E2)/P(E1)=n(E2)/n(E1)=4/10=2/5
• 11 years agoHelpfull: Yes(32) No(3)
• 5 = (1,4),(2,3),(3,2),(4,1)
  7 = (1,6),(2,5),(3,4),(4,3),(5,2),(6,1)
  n(S)= 4+6=10
  n(E)= 4
  p(E)= 4/10 = 2/5
• 11 years agoHelpfull: Yes(8) No(4)
• Let A=P(sum=5)=4/36 {(1,4) (4,1) (2,3) (3,2)}
  B=P(sum=7)=6/36 {(6,1) (1,6) (2,5) (5,2) (3,4) (4,3)}
  C=P(sum other than 5 or 7) 1-(10/36)=26/36
  P(sum 5 comes b4 7)=P(5)+P(C)P(5)+P(C)P(C)P(5)....
  =4/36(1+26/36+(26/36)^2+(26/36)^3......)
  =4/36*1/(1-26/36)=(4/36)*(36/10)=2/5
  thus ans=2/5
• 11 years agoHelpfull: Yes(2) No(1)
• total ways of getting =4, total ways of getting 7=6
  
  so, Probability of 5 before 7 = 4/10 Ans
• 11 years agoHelpfull: Yes(1) No(0)