on a toss a two dice.a throw a total of '5' then the probability that he will throw another '5' before he throws '7' is?

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