Am I correctly using the conditional probability notation in this problem? Problem: Two numbers are chose at random from among the numbers 1 to 10 without replacement.
Goal: Find the probability that the second number chosen is 5.
Official Solution:
$A_{i}:i \in \left\{1,2,...,10\right\}$ is event that first number chosen is $i$. $B$ is the event that second number chosen is 5. Use the total probability law to get: $$\begin{equation}P(B)=\sum_{i=1}^{10}P(B|A_{i})P(A_{i}) = 9 \, \frac{1}{9} \, \frac{1}{10}=\frac{1}{10}\end{equation}$$
My Solution:$$\begin{equation*}P(B|A_{i}) = \frac{n(B)}{n(S)} = \frac{9}{9 \cdot 10} = \frac{1}{10}\end{equation*}$$
Question: How can we at all write $P(B)$? Is probability for event $B$ not inherently conditional? You can't choose the second number without choosing the first number, so how could one write anything other than $P(B|A)$?

Alternative approach:

Question: How can we at all write $P(B)$? Is probability for event $B$ not inherently conditional? You can't choose the second number without choosing the first number, so how could one write anything other than $P(B|A)$?

First thought experiment: 
Lottery with tickets 1 through 10.  A mediator picks two of the lottery tickets.  He then arbitrarily chooses one of the two tickets and places it in Envelope-1.  He then places the other ticket in Envelope-2.  Clearly, the chance of the ticket-5 being in Envelope-1 is the same as the chance of the ticket-5 being in Envelope-2.
Second thought experiment: 
Same as the first, but instead, the mediator picks the two numbers one at a time.  The first number picked goes in Envelope-1.  The second number picked goes in Envelope-2.  The analysis is identical to the first thought experiment.  It doesn't matter whether Envelope-1 contains the first ticket (re second thought experiment) or whether it is 50-50 whether Envelope-1 contains the first ticket or the second ticket (re first thought experiment).
Therefore, the problem can be reduced to the probability of drawing a 5 on the first draw.  Therefore, without involving any conditional probabilities, $p(B) = \frac{1}{10}.$
However, the only way to utilize the above approach is to first develop your intuition to the point that you are absolutely certain that my approach is valid.
Addendum
Justifying the analysis from a different perspective.
The tickets 1 thru 10 are lined up in a row.  Then the mediator randomly assigns a number 1 thru 10 (without replacement) to each ticket.  The assigned number represents when the ticket will be drawn.
That is, if the ticket-5 is assigned the number 1, it will be drawn first.  If the ticket-5 is assigned the number 2, it will be drawn second.
The ticket-5 is equally likely to be assigned the number 1 as it is to be assigned the number 2.  Therefore, from this alternate perspective, (again) the chance of ticket-5 being drawn second is the same as the chance of ticket-5 being drawn first.
