A & B throws a pair of dice and winner is one who gets a sum of 9 first.if A starts throwing first
Then what is the ratio of A winning over B?

• scince two dice are thrown together
  therefor total no of ways = 6*6 = 36
  science they wants 9 thatsby that may be (3,6) or (6,3) or (5,4) or (4,5)
  total no of ways for winning = 4
  probablity of winning of A = 4/36 = 1/9
  science A starts the game then B will only win when A will lose
  probablity of lossing of A = 32/36 = 8/9
  therefor probablity of winning of B
  =prbablity of lossing A*probablity of winning of B when B starts the game
  = 32/36*4/36
  = 8/81
  
  Required ratio = (1/9)/(8/81) = 9 : 8
• 8 years agoHelpfull: Yes(18) No(1)
• probability of a person throwing first and winning is given by 1/(2-p)
  1/(2-(1/9))=9/17 prob of a
  prob of b is 8/17
  prob (a/b) 9/8
• 8 years agoHelpfull: Yes(3) No(9)
• total no of ways =36
  probability of A to win =
  there are 4 cases if he gets (3,6) (6,3) (5,4) (4,5)
  probability of winning of A =4/36
  =1/9
  probability of losing=1-(1/9)
  =8/9 =probability of winning of B
  REquired ratio=1/9():(8/9)
  =1:8
• 7 years agoHelpfull: Yes(0) No(4)
• 1/9 + ( 8/9 x 8/9 x 1/9 ) + ( 8/9 x 8/9 x 8/9 x 8/9 x 1/9 ) + ......
  1/9 [ 1 + (8/9)^2 + (8/9)^4 + (8/9)^6 +....infinite G.P series ]
  1/9 x [ 1/(1 - (8/9)^2)]
  9/17
• 4 years agoHelpfull: Yes(0) No(0)