Digits $=3,3,4,4,4,5,5$

Total $7$ digit numbers $=\frac{7!}{2!2!3!}$

Number of $7$ digit number divisible by $2$ 

$\Rightarrow$last digit $=4$

Now $7$ digit numbers which are divisible by $2$

$=\frac{6!}{2!2!2!}$
Required probability $=\frac{\frac{6!}{2!2!2!}}{{\displaystyle \frac{7!}{3!2!2!}}}=\frac{3}{7}$