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 $P(B|A)$?

You are asked to evaluate the probability that the second number chosen is $5$. This is clearly $P(B)$. There is no condition mentioned.
Yes you are right that you cannot choose a second number without having chosen the first number. But $P(B)$ this is the probability that the second chosen number is $5$ regardless which number is chosen first. You must choose a first number, but it is irrelevant which one.

For this purpose the law of total probabilty is used. For instance
$P(B|A_1)$ is the probability that the second number is a $5$ given that the first number is $1$.
The number $1$ has been already drawn first. 9 numbers still can be secelcted. Number 5 is still among these numbers. Thus  $P(B|A_1)=\frac19$
$P(A_1)$ is the probability that number 1 is selected first is $\frac1{10}$.
Thus $P(B\cap A_1)=P(B|A_1)\cdot P(A_1)=\frac{1}{90}$
The probability $P(B\cap A_i)$ are all $\frac{1}{90}$ except for $i=5$. $P(B\cap A_5)=0$
Thus $P(B)=\sum\limits_{i=1}^{10} P(B|A_i)\cdot P(A_i)$
$=\frac1{90}+\frac1{90}+\frac1{90}+\frac1{90}+0+\frac1{90}+\frac1{90}+\frac1{90}+\frac1{90}+\frac1{90}+\frac1{90}=\frac{9}{90}$

P(B) can be calculated more easily.  To draw a 5 secondly it must not be drawn at the first draw. $P(\overline A_5)=\frac9{10}$.
$\overline A_5$ is the complementary event of $A_5$. This it the event that $5$ is not chosen first. And $P(B|\overline A_5)=\frac19$. This is the probabilty that number 5 is chosen secondly with the condition that 5  is not chosen first. Using the total law of probability:
$P(B)=(B|\overline A_5)\cdot P(\overline A_5)+(B| A_5)\cdot P( A_5)=\frac19\cdot  \frac{9}{10}+0\cdot \frac1{10}=\frac1{10}$
I cannot really comprehend what do you mean by $n(B)$ and $n(S)$. And how you got the numbers for them ?
