Prove the following: What is the probability that a randomly chosen 4 digit integer has distinct digits and is odd?

Let us consider the digits in the order last - first-second-third

There are 5 ways of selecting the last digit (1,3,5,7,9)
There are 8 ways of selecting the first digit since a FOUR DIGIT INTEGER can't start with 0
There are no restrictions on choosing from the remaining 8 digits,
so favorable ways = 5*8*8*7
total ways = 9000 (1000-9999)
indicated Pr = 5*8*8*7/9000 = 56/225, approx 0.2489 <------

The first digit can be any digit (I suspect that 0 is not allowed as a first digit)
9 out of nine = 100% = 1

The second digit can be any digit, except the one picked for the first digit. However, 0 is now allwed.
9 out of 10 = 0.9

The third digit can be any of the remaining digits
0.8

The last digit must be an odd digit (1/2 probability)
0.5

Total probability= multiplication of the 4 probabilities:
(1)(0.9)(0.8)(0.5) = 0.36

There's only one kind of digit, and the probability is 50% since there's the same number of odd numbers as there is even