The probability of getting a total of 5 using a pair of dice is $ \large\frac{1}{9}$ and that
of getting a total of 7 is $ \large\frac{1}{6}$
Also $ P( \overline 5\: \overline 7) = \large\frac{36-10}{36}$
$ \large\frac{13}{18}$
Because there are 10 cases in favour of getting a sum of 5 or 7.
5 can come before 7 in the folowing cases : 5 or $ ( \overline 5\: \: \overline 7 ) $
5 0r $ ( \overline 5\: \: \overline 7 ) \: ( \overline 5\: \: \overline 7 ) $ 5 or.......
Hence the required probability is
$ \large\frac{1}{9} + \large\frac{13}{18} \times \large\frac{1}{9} + \large\frac{13}{18} \times \large\frac{13}{18} \times \large\frac{1}{9} +.....$
$ \Large\frac{ \Large\frac{1}{9}}{1-\Large\frac{13}{18}}$
$ = \large\frac{2}{5}$
Ans : (A)