Three dice are rolled. Find the number of possible outcomes in which at least one dice shows $5$.
I would like to know the error in my attempt, I am aware of the other method which is $6^3-5^3$.
Attempt:
Let the first die show 5. The possibilities for the other two dice are $6\times 6 =36$.
Same can be done with the other two dice.
As a result, answer should be $36 \times 3 - 2 = 108 - 2 = 106$, ($-2$ because of the repetition of 5,5,5 case $3$ times. )
On
Your method has the looks of a wrong application of PIE.
The correct one is:$$|E_1\cup E_2\cup E_3|=|E_1|+|E_2|+|E_3|-|E_1\cap E_2|-|E_1\cap E_3|-|E_2\cap E_3|+|E_1\cap E_2\cap E_3|=$$$$6^2+6^2+6^2-6-6-6+1=91$$
Here $E_i$ denotes the event that the $i$-th die shows a $5$. This for $i=1,2,3$
Also there is the more simple approach (see my comment on your question) leading to answer $6^3-5^3=91$ and not using PIE.
It seems you made it:$$|E_1\cup E_2\cup E_3|=|E_1|+|E_2|+|E_3|-3|E_1\cap E_2\cap E_3|$$which is wrong.
Total Number of elements is sample space =$6^3=216$.
Total number of elements in the sample space which contains no $5$ = $5^3=125$.
Thus total number of elements which having at least one $5$ is $=216-125=91$.
Update after OP request:
If there is only one $5$. Total Number of possibilities=$1×5×5×3C1=75$.
If there is only two $5$. Total Number of possibilities=$1×1×5×3C2=15$.
If there is only three $5$. Total Number of possibilities=$1×1×1=1$.
Total$=75+15+1=91$