### Solution:

It is given that boxes contain some slips with Re.1, Rs.5, and Rs.13.We have to find the probability that the marked slip is other than $\n \n Re.1\n $.

First we find the total number of slips.

$\n \n n(S)=25+50\n $

$\n \n =75\n $

$\n \n 19\n $ are marked in Box $\n A\n $ and $\n \n 45\n $ are marked in Box $\n B\n $ as Re.1.

So, $\n \n 19+45\n $ $\n \n =64\n $.

The number of slips that are marked other than $\n \n Re.1\n $ is,

$\n \n 75\u221264=11\n $.

The probability of slips that are marked other than $\n \n Re.1\n $ is,

$\n \n =\n \n 11\n \n 75\n \n \n $.

The probability that slip is marked other than is $\n \n \n \n 11\n \n 75\n \n \n $.

Hence, option 2) is correct.