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# A pair of dice is rolled together till a sum of either 5 or 7 is obtained. The probability that 5 comes before 7 is

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

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A)
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)