Four bars are drawn without replacement from ten bars marked from $1$ to $10$. Find the probability of smallest number being $5$. A bag contains ten bars marked from $1$ to $10$. Four bars are drawn without replacement, the number noted. Find the probability of:
i. The smallest number is $5$.
ii. The biggest number is $5$.
My Attempt:
i. For the smallest number $5$ $\to$ favorable cases  $5,6,7,8,9,10$ (unordered)
total probability space: $\binom{10}{4}$
Pr(of being smallest number 5): $\dfrac{\binom{6}{4}}{\binom{10}{4}}$
ii. for the biggest number $5$ $\to$ favourable cases is $1,2,3,4,5$ (unordered)
total probability space: $\binom{10}{4}$
pr(of being Biggest number 5): $\dfrac{\binom{5}{4}}{\binom{10}{4}}$
I am confused about how do I remove the other possibilities in both cases? For example, the smallest number $5$, how do I exclude the probability of not having $1,2,3,4$?
 A: Note that in order for the smallest number to be $5$, you must draw the $5$ and any three of the numbers $6,7,8,9$, and $10$, which can be done in $\binom53=10$ ways. Similarly, in order for the largest number to be $5$, you must draw the $5$ and any three of the numbers $1,2,3$, and $4$, which can be done in $\binom43=4$ ways.
You can start with the calculations that you made and subtract the unwanted cases, but that approach is more work than is really necessary. In the first problem, for instance, you get $\binom64$ ways to draw four cards from the set $\{5,6,7,8,9,10\}$. You don’t want the sets of four cards drawn from the set $\{6,7,8,9,10\}$, and there are $\binom54$ of those, so you really have just $\binom64-\binom54=15-5=10$ possibilities. Similarly, in the second program you want to discard the draw of $1,2,3$ and $4$, so instead of $\binom54=5$ possibilities, you have only $\binom54-1=4$ possibilities.
A: What you have computed in i is the chance that the smallest is at least $5$, not that it is exactly $5$ because you have computed the chance that all four are in the range $5$ to $10$.  You need to compute the number of ways you can get $5$ and three out of $6-10$, which is $\frac {1C1 \cdot 5C3}{10C4}$.  
You have a similar problem with ii.
