Let X be the set of all five digit numbers formed using 1, 2, 2, 2,...

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,

$\mathop{- - - - 0}_{⏟}$

Now in first four place following number can take place,

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

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

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

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

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

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

$\mathop{- - - 1 0}_{⏟}$

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, $p = \dfrac{31}{38}$

$⇒ 38p = 31$