Let X be the set of all five digit numbers formed using 1 , 2 , 2 , 2 , 4 , 4 , 0 . For example, 22240 is in…

First we will find the sample space in which the number of five-digit numbers are divisible by $5$,

So, fixing zero at the last place we get,

$\underbrace{----\underline{0}}$

Now in first four place following number can take place,

$2224 \to \frac{4!}{3!}=4 \text{ways}$

$2244\to \frac{4!}{2!2!}=6 \text{ways}$

$2221\to \frac{4!}{3!}=4 \text{ways}$

$2241\to \frac{4!}{2!}=12 \text{ways}$

$2441\to \frac{4!}{2!}=12 \text{ways}$

So, total sample space will be $4+6+4+12+12=38$ 

Now finding the number of favourable outcomes,

So, Number of five-digit numbers divisible by $5$ but 'not' by $20$

Now fixing $10$ in last two places, we get

$\underbrace{---\underline{1} \underline{0}}$

So, the first three places can be occupied by,

$222 \to 1 \text{ways}$

$224 \to 3 \text{ways}$

$244 \to 3 \text{ways}$

So, total number of numbers which are divisible by $5$ but not $20$ will be, $1+3+3=7$

So, favourable number of five-digit numbers that are divisible by $5$ and $20=38-7=31$

Hence, probability is given by, $p=\frac{31}{38}$

$\Rightarrow 38 p=31$