Box-I contains 3 cards bearing numbers 1 , 2 , 3 ; Box II contains 5 cards bearing numbers 1 , 2 , 3 , 4 , 5…

Given:-

Box-I contains  $3$ cards bearing numbers $1,2,3.$

$\therefore$Probability of selection of even no. from box-I $=\frac{1}{3}$

& Probability of selection of odd no. $=\frac{2}{3}$

Box-II contains $5$cards bearing numbers $1,2,3,4,5$

$\therefore$Probability of selection of even no. from box-II $=\frac{2}{5}$

and probability of selection of odd no. $=\frac{3}{5}$

Box-III contains $7$ cards bearing the numbers $1,2,3,4,5,6,7$.

$\therefore$Probability of selection even no. from Box-III $=\frac{3}{7}$

and probability of selection of odd no. $=\frac{4}{7}$

Now,

Required probability $=\text{sum of numbers}\left(x_1+x_2+x_3\right)$selected from all the three boxes is odd.

$=P\left(\text{when all the three cards are odd}\right)+P\left(\text{when two are even \& one is odd}\right)$

$=\left(\frac{2}{3}\times \frac{3}{5}\times \frac{4}{7}\right)+\left(\frac{1}{3}\times \frac{2}{5}\times \frac{4}{7}\right)+\left(\frac{1}{3}\times \frac{3}{5}\times \frac{3}{7}\right)+\left(\frac{2}{3}\times \frac{2}{5}\times \frac{3}{7}\right)$

$=\frac{53}{105}$