Correct Answer - B
5 can be thrown in 4 ways and 7 can be thrown in 6 ways, hence number of ways of throwing neither 5 nor 7 is `36-(4+6)=26`
`:." Probability of throwing a five in a single throw with a pair of dice `=(4)/(36)=(1)/(9)` and probability of throwing neither 5 nor `7=(26)/(36)=(13)/(18)`
Hence, required probability
`=((1)/(9))+((13)/(18))((1)/(9))+((13)/(18))^(2)((1)/(9))+...=((1)/(9))/(1-(13)/(18))=(2)/(5)`