<s>[INST] Four numbers drawn from 1 to 100000 randomly, the same number may be chosen more than once. Four numbers drawn from 1 to 100000 randomly, the same number may be chosen more than once. Determine probability the last digit from multiplying that four numbers is 1 or 9.
I have tried with many cases with it but it come a big number on combination. 
 [/INST] As mentioned in the comments, we may assume we only draw numbers from the set $\{0,1,2,\ldots,9\}$. This is the case because only the last digit of the number we picked matters. Let us call the random numbers $A$, $B$, $C$, $D$. If one of $A$, $B$, $C$, $D$ is even or divisible by $5$, then $ABCD$ is as well and then it will not end in a $1$ or $9$. The probability that $ A$, $B$, $C$, $D$ are all coprime to $10$ is $(\frac25)^4$ (we can only pick $1$, $3$, $7$, $9$). We can now show that the last digit of $AB$ takes these values $1$, $3$, $7$, $9$ with equal probability and that the same holds for $ABC$ and $ABCD$. In half of the cases the last digit will be $1$ or $9$. This yields a final answer of 
$$
\left(\frac25\right)^4 \cdot \frac12 = 0.0128.
$$
Edit: Consider the multiplication table of $1$, $3$, $7$ and $9$ (we consider last digits only):
$$
\begin{array}{c|cccc}
\times & 1 & 3 & 7 & 9 \\ \hline
1 & {\color{green}1} & 3 & 7 & {\color{green}9} \\
3 & 3 & {\color{green}9} & {\color{green}1} & 7 \\
7 & 7 & {\color{green}1} & {\color{green}9} & 3 \\
9 & {\color{green}9} & 7 & 3 & {\color{green}1}
\end{array}
$$
Exactly half of the entries of this table is a $1$ or a $9$, so if the last digits of $X$ and $Y$ are independently equidistributed on $\{1,3,7,9\}$ then so is the last digit of $XY$.
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