Question

What is probability that from all 5-digits numbers that have distinct digits randomly chosen number will be odd? Divisible by 5?

Answer 1

I understand that there are 9*9*8*7*6 such 5-digit numbers. However, how do I find how many of them are odd?

Answer 2

It would be easy, if only last digit could be an odd digit, but all of them can and that is the obstacle for me.

Answer 3

okay for odd, we can divide the digits in places
Like this _ _ _ _ _
For 5 digits distinct number, the five places can be occupied by a number from 1 till 9
But for an odd number the last place can only be occupied by either 1,3,5,7 or 9
Agree?

Answer 4

I definitely agree with that ;) But I can't write 9 * 9 * 8 * 7 * 5 since it is possible that one of the odd digits were used somewhere in the number and the units place would get only 4 or 3 or 2 or 1 digit to choose from. How do I take into account all of such cases?

Answer 5

Okay so, we know that only 5 digits out of those 9 digits can end up in the last place
But for every case, only one (out of those 5 ) will end up in the last place.
As a result we have to choose the remaining four places from those 8 digits
Agree?

Answer 6

Hm, I got the first part, but am I wrong or you aren't taking into account 0?

Answer 7

Because, shouldn't it be 8 x 8 x 7 x 6 x 5? Since first digit can't be zero but all others apart from the last can?

Answer 8

First digit = 8 numbers (excluding 0 and last digit number)
Second digit = 8 numbers (excluding first and last digit number)
Third digit = 7 numbers (excluding 1st,2nd, last digit number)
Fourth digit = 6 numbers (excluding 1st,2nd, 3rd, last digit number)
Fifth digit = 5 numbers (only odd numbers from 0 till 9)

Answer 9

Checked the answer and it is right! Thank you :)

As for the second part of the question (about number of divisible by 5)
Similar method: 8 x 8 x 7 x 6 x 2, right?

Answer 10

(I think that it somehow is not right with numbers divisible by 5 because the answer in my book is different, but that might be the book's mistake)

Answer 11

No maybe book is right
we have to consider two cases.
One ending with 5 and one ending with 0. (Can you figure out why we are treating these two as separate cases?)

Answer 12

Oh. I think, in one case the number will end with 0, so first digit can be anything out of 9 digits. In another one the number will end with 5, so the first digit could be one out of 9. Am I right?

Answer 13

Solved it when I considered both cases separately, thank you for your help @FaiqRaees ! :)

Answer 14

See the problem was when the last number is 0, the other digits cannot use 0 as a digit
But when we consider 5, the first digit cannot use 0 but others can. The limitation in the values of first digit is what making us to treat these two cases separately