A and B throws dice.what is the probability that A's throws is not greater than B's throws?

• A's throw throw could be greater than B by following ways
  
  (2,1)
  (3,2) (3,1)
  (4,3) (4,2) (4,1)
  (5,4) (5,3) (5,2) (5,1)
  (6,5) (6,4), (6,3) (6,2), (6,1)
  
  So in total, there are 15 possible outcomes in which the throw could exceed.
  
  So in the probability in which A's throw can exceed is
  
  15/36 or 5/12
  
  Probability of failure = 1-p
  
  = 1-15/36 = 21/36 = 7/12
  probability that A's throws is not greater than B's throws=7/12(ans)
  
• 12 years agoHelpfull: Yes(52) No(1)
• this will include the cases when A and B get the same value
  so the cases possible are:
  (1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
  (2,2) (2,3) (2,4) (2,5) (2,6)
  (3,3) (3,4) (3,5) (3,6)
  (4,4) (4,5) (4,6)
  (5,5) (5,6)
  (6,6)
  so the probability is 21/36 =7/12
• 12 years agoHelpfull: Yes(12) No(1)
• 7/12 is the right answer.
  
• 12 years agoHelpfull: Yes(2) No(0)
• A is not greater than B which means that it can be either equal or greater. so the probability can be . . . .
  
  when A throws 1 : B can throw only 1 => 1 possibility
  when A throws 2 : B can throw only 1,2 => 2 possibility
  when A throws 3 : B can throw only 1,2,3 => 3 possibility
  when A throws 4 : B can throw only 1,2,3,4 => 4 possibility
  when A throws 5 : B can throw only 1,2,3,4,5 => 5 possibility
  when A throws 6 : B can throw only 1,2,3,4,5,6 => 6 possibility
  hence the required possibility is (1+2+3+4+5+6)/36(total possibility) = 7/12
  
• 11 years agoHelpfull: Yes(1) No(0)
• A=1,2,3,4,5
  nd B=6
  so prob.=(6c5.6c1)/6.6=1
• 12 years agoHelpfull: Yes(0) No(12)
• ans - 7/12
  
  not greater than => less than or equal to => 21 outcomes
  
  probability = 21/36 = 7/12
• 12 years agoHelpfull: Yes(0) No(0)
• answer =5/12
• 9 years agoHelpfull: Yes(0) No(0)
• B 2 3 4 5 6
  
  A 1 1,2 1,2,3 1,2,3,4 1,2,3,4,5
  
  total possibilities are: 15
  total probability: 1
  no of throws : 36
  probability = 1-(15/36)
  = 7/12
• 9 years agoHelpfull: Yes(0) No(0)