$\begin {array} {1 1} (A)\;\large\frac{2}{5} & \quad (B)\;\large\frac{1}{5} \\ (C)\;\large\frac{3}{5} & \quad (D)\;None\: of \: these \end {array}$

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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)

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