numbers in a 5 digit number i have a very simple question 
i need to know the probability of a 5 digit number to be with the digit 5 only one time 
so first digit cant be 0 so i do:
$8\times9\times9\times9\times1$
$8\times9\times9\times1\times9$
$8\times9\times1\times9\times9$
$8\times1\times9\times9\times9$
that is:
$8\times9\times9\times9\times4$
but if the first number is five it's
$1\times9\times9\times9\times9$ and it does not come to:
$8\times9\times9\times9\times5$ because now you can have 0 in the remain digit..
how do i calculate it?
 A: The total amount of $5$-digit numbers is $9\cdot10\cdot10\cdot10\cdot10=90000$.
The amount of those with $5$ only at index #$1$ is $1\cdot9\cdot9\cdot9\cdot9=6561$.
The amount of those with $5$ only at index #$2$ is $8\cdot1\cdot9\cdot9\cdot9=5832$.
The amount of those with $5$ only at index #$3$ is $8\cdot9\cdot1\cdot9\cdot9=5832$.
The amount of those with $5$ only at index #$4$ is $8\cdot9\cdot9\cdot1\cdot9=5832$.
The amount of those with $5$ only at index #$5$ is $8\cdot9\cdot9\cdot9\cdot1=5832$.
So the probability of a $5$-digit number with only one occurrence of the digit $5$ is:
$$\frac{6561+5832+5832+5832+5832}{90000}=0.3321$$
A: First, you are counting the number of five digit numbers that include exactly one five, not computing the probability.  For the probablility you need to divide by the number of five digit numbers, which is $9\cdot 10^4$  Your counts are correct, so the number we want is $8 \cdot 9 \cdot 9 \cdot 9 \cdot 4+1 \cdot 9 \cdot 9 \cdot 9 \cdot 9$  You could just type this into your calculator (you may need to use the memory to hold the first term) or you can write  $8 \cdot 9 \cdot 9 \cdot 9 \cdot 4+1 \cdot 9 \cdot 9 \cdot 9 \cdot 9=(8\cdot 4+1\cdot 9)9^3=41\cdot 729=29889$
A: One digit is 5, so there are 4 free ones left. The first digit has 8 possibilities (all digits except 5 and 0), the second, third and fourth all have 9 possibilities (all digits except 5). This makes for $8\times9\times9\times9=5832$ possibilities. However, the 5 could be in any position. There are 5 possible positions for the digit 5.
Four of these do not have the 5 as the first digit. One does, for a total of $4\times5832+9^4=29889$ possible numbers.
The total amount of possible 5 digit numbers without a leading zero is $9\times10^4$. The chance is therefore $\frac{29889}{9\times10^4}=0.3321$
A: If the first number is 5 then it's $1\times9\times9\times9\times9$.
If not then it's $8\times1\times9\times9\times9$ or $8\times9\times1\times9\times9$... which is $4\times8\times1\times9\times9\times9$.
So the number of cases is $4\times8\times1\times9\times9\times9+1\times9\times9\times9\times9=29889$
A: Since the first digit canot be a zero, the probability that the first digit is a 5 is given by:$$\frac{1}{9}$$And the probability that the remaining four digits of this number are not 5's is given by:$$\left(\frac{9}{10}\right)^4$$So the probability that the first digit is a 5 and the remaining digits are not 5's is given by:$$\frac{1}{9}\times\left(\frac{9}{10}\right)^4$$Next we consider the digit 5 appearing as the second digit of the number. The probability that the second digit is a 5 is given by:$$\frac{1}{10}$$The probability that the first digit of this number is not a 5 (remember it cannot be a zero either) is given by:$$\frac{8}{10}$$The probability that the remaining three digits of this number are not 5's is given by:$$\left(\frac{9}{10}\right)^3$$Therefore, the probability that the second digit of a 5 digit number is a 5 is given by:$$\frac{1}{10}\times\frac{8}{10}\times\left(\frac{9}{10}\right)^3$$Hopefully you can complete the caculation from here...
A: In combinatorics:
There are (9^^4 )× 1 = 6561 
+
(9^^3) × (8/9) = 648
= 7209 ways to have  5 in on  and only one column 
10^^4 × 9 
= 90000 ways to order five significant digit numbers
.: there is  a probability  of 7209 / 90000
= .00801 probability of having a five in a 5 digit number
A: Mr Milican is in part correct. I will  do  some recalculation.
Firstly if 5 is in the first  column then  there is the instance of  1 possible  number with  the probability of being  5 at exactly  1.  Now we have four digits that must not be 5 in any digit. There are 9 possibilities of digits  not  being  5 in each of these digits.  Hence 9 ^^ 4  is indeed the  cimbinatoric  number of permutations of any non 5 digit combination that can be made in any instace.
= 6561 combinatiins
Secondly if the first digit is not a 0 or a 5 then there are 8 possible other digits that the first and highest  magnitude digit may  be.  In that case  there are  4 possibilities of a 5 being in one and only  one  digit of the 4 lowest  magnitudes.  And  there 3 digits where it must not  be  a 5. In each of these magnitudes there  are  9 possible digits such that in a combinatoric computation the total number  of combinations is( (9^^3) × 4  ) × 8 =
23328
The sum total of possibilities  is
23328 + 6561
= 29888
And since there are 9 × (10^^4)
= 90000 possible 5 digit  numbers 
The probability  a 5 digit  number has 5 in one and only one digit is 
29889 / 90000
= 0.3321
Fin 
