on a toss of two dice A throws a total of 5. what is the probability that he throws another 5 before he throws 7

• 5 can be thrown in 4 ways and 7 in 6 ways in a single throw with a pair of dice. Hence number of ways of throwing neither 5 nor 7 is 36 - ( 4 + 6 ) = 26
  
  Hence probability of throwing a 5 in a single throw with a pair of dice is 4/36 = 1/9
  
  And probabilty of throwing neither 5 nor 7 is ( 26/36 ) = ( 13/18 )
  
  Hence the required probability
  
  = ( 1/9 ) + ( 13/18 ) ( 1/9 ) + ( 13/18 )² ( 1/9 ) + ( 13/18 )³ ( 1/9 ) + .....
  
  = ( 1/9 ) / [ 1 - ( 13/18 ) ] = 2/5
  
  EXPLAINATION :
  
  Since he has already thrown a 5, ( ie a number different from 7 ) , he may throw 5 at the next attempt. The probabilty for which is ( 1/9 ) or he may throw 5 at the second attempt when he fails to throw either 5 or 7 at the first attempt, the probability for which is ( 13/18 ) ( 1/9 ) or at the third attempt the probability for which is ( 13/18 )² ( 1/9 ) and so on ........ Adding all the probabilities we get the answer.
• 10 years agoHelpfull: Yes(15) No(2)
• I think it could be 1/9
  because probability to getting a sum of 5 is (1,4),(4,1),(3,2),(2,3)
  S=36
  now p=4/36=>1/9
• 10 years agoHelpfull: Yes(0) No(25)
• ans 4/36
  total exp r 36
• 10 years agoHelpfull: Yes(0) No(8)
• can any1 explain me clearly.
• 5 years agoHelpfull: Yes(0) No(0)