what is the probability of getting 5 before 7 while 2 dice r thrown?

• p(5)={(1,4);(2,3);(3,2);(4,1)}=4
  p(7)={(1,6);(2,5);(3,4),(4,3);(5,2);(6,1)}=6
  so ans=4/(4+6)
  =0.4
• 10 years agoHelpfull: Yes(24) No(0)
• my ans is 2/5
• 10 years agoHelpfull: Yes(7) No(1)
• p=2/5;
  p(5)=4;
  p(7)=6;
  n(s)=10;
  
• 10 years agoHelpfull: Yes(4) No(1)
• probability is 11/36.(total no of events 36)
• 10 years agoHelpfull: Yes(3) No(7)
• N(e)=p(5)=4
  P(7)=6
  N(s)=6+4=10
  P(e)=n(e)/n(s)
  =4/10=2/5
• 10 years agoHelpfull: Yes(3) No(0)
• its 1/9 as 5 can come from (1,4),(2,3),(4,1),(3,2). so the probability will be 4/36
• 10 years agoHelpfull: Yes(2) No(1)
• probability of getting 5 is 4/36
  probability of getting 7 is 6/36
  
  getting 5 before 7 is (4/36)+(6/36 * 4/36)+(6/36 * 6/36 *4/36)......
  it is an infinite gp...
  so sum is 2/15
  
• 10 years agoHelpfull: Yes(2) No(0)
• 12/36=1/3
• 10 years agoHelpfull: Yes(0) No(1)
• pls explain
  
• 10 years agoHelpfull: Yes(0) No(0)
• Can anyone tell where in the question sum of 5 or 7 is written??
• 10 years agoHelpfull: Yes(0) No(0)
• 4/10
• 10 years agoHelpfull: Yes(0) No(0)
• 2/5 is the answer
  
• 10 years agoHelpfull: Yes(0) No(0)
• probability for getting 5 = 4/36
  probability for getting 7 = 6/36
  so the probability of getting 5 before 7 = ((4/36))/(4/36)+(6/36)
  ans= 2/5
  
• 10 years agoHelpfull: Yes(0) No(0)