Answer:
(a) 
(b) 
(c) 
(d) 
(e) 
--- Probability of distinct digits
--- Probability of odd distinct digits
Step-by-step explanation:
Solving (a): Integers from 1000 to 9999
To do this, we simply add 1 to the range.
i.e.
Range is the difference between the given interval.
Solving (b): Odd integers
This implies that the last digit must be any of 1, 3, 5, 7 and 9 (i.e. 5 digits)
The first digit can not be 0 (i.e any of the remaining 9 digits)
There is no restriction to other digits
Numbers from 1000 to 9999 are 4 digits, so the possible selection are:
Total selection is:
Solving (c): Distinct digits
This implies that all 4 digits are different and the first can not be 0.
So, we have:
i.e. (1 - 9)
i.e. (0 - 9) minus the first digit
Total selection is:
Solving (d): Odd digits that are distinct
This implies that the last digit must be any of 1, 3, 5, 7 and 9 (i.e. 5 digits)
The first digit can not be 0 and must be different from the last (i.e 8 digits)
The second digit must be different from the first and the last(i.e. 8 digits)
The third digit must be different from the three other digits (i.e. 7 digits)
So,
Solving (e): Probability that a number is distinct
In (a), total possible digits is 9000
In (c), total distints are 4536
So, the probability is:
In (d), total odd distinct digits are 2240
So, the probability is:
