Not divisible by 5

Let $P_A$ be the probability that the numbers formed are divisible by $5$ . Then the probability that the numbers formed are not divisible by $5$ is $1-P_A$ .

If the last digit is $5$ , there are $3 \times 3 \times 2 \times 1 \times 1 = 18$ numbers.

If the last digit is $0$ , there are $4 \times 3 \times 2 \times 1 \times 1 = 24$ numbers.

The total is $18+24=42$ numbers.

The total numbers that can be formed is $4 \times 4 \times \times 3 \times 2 \times 1=96$ numbers.

So $P_A=\dfrac{42}{96}=0.4375$ . Finally, $1-P_A=1-0.4375=$ $\color{#D61F06}\boxed{\large 0.5625}$

We know that a number is divisible by 5 when it has a 0 or a 5 in the unit place. Keeping 5 in the unit place, we arrange the 4 numbers in 4!=24 ways. Numbers that have 0 in the front are meaningless. Keeping 0 in the front, we arrange the 3 numbers in 3!= 6 ways. Numbers that have 5 in the unit place are 24-6=18 numbers. Numbers that have 0 in the unit place can be arranged in 4!=24 ways. Numbers that are divisible by 5 are 24+18=42 numbers. There are 5!=120 numbers that can be formed with all the 5 numbers. Numbers that are meaningless have 0 in the front and they are 4!=24. 120-24=96 numbers. The probability of getting numbers that are divisible by 5 is 42/96= 0.4375 and that of not getting is 0.5625.