We have a bag which contains $49$ numbers. We draw $6$ numbers randomly without replacement.
What is the probability that the fourth number is greater that the first one?
My approach:
First we define the sample space by $$\Omega:=\{(\omega_1,\dots,\omega_6)\mid 1\leq\omega_i\leq 49,~ \omega_i\neq\omega_j \text{ where } 1\leq i\neq j\leq 6\}$$ and the two sets $$A_1:=\{(\omega_1,\dots\omega_6)\mid \omega_1>\omega_4\}\text{ and } A_4:=\{(\omega_1,\dots\omega_6)\mid \omega_1<\omega_4\}.$$ We are looking for $\mathbb{P}(A_4)$.
As each outcome has the same probability we have $\mathbb{P}(A_4)=\frac{|A_4|}{|\Omega|}$, where $|A|$ denotes the cardinality of a set $A$. It is clear that $A_1$ and $A_2$ are disjoint and $\Omega=A_1\cup A_4$.
We define the function $f:A_1\to A_4$, where \begin{align*} &f(\omega)=b\implies f((\omega_1,\dots\omega_6))=\begin{cases}b_1=\omega_4\\b_2=\omega_2\\b_3=\omega_3\\ b_4=\omega_1\\b_5=\omega_5\\b_6=\omega_6.\end{cases} \end{align*} It is obvious that $f$ maps to $A_4$, is surjective and injective. So $f$ is a bijection and $|A_1|=|A_4|$. Hence, $\mathbb{P}(A_4)=\frac{|A_4|}{|\Omega|}=\frac{\frac{1}{2}|\Omega|}{|\Omega|}=\frac{1}{2}$.
Is this correct?