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Your answer is correct !

Here is just another way looking at only the two non-aces and the four aces, with $N$ denoting non-aces, and $A$ denoting aces.

$\boxed{N}\boxed{N}\quad\boxed{A}\quad\quad\boxed{A}\quad\quad\quad\boxed{A}\quad\boxed{A}\quad Pr = \dfrac{48}{52}\dfrac{47}{51}\dfrac{10}{50}\dfrac9{49}\dfrac{8}{48}\dfrac7{47}$

The idea is that in sequence drawn, the first two must be non-aces, and there must be four aces somewhere in the next ten.

Your answer is correct !

Here is just another way looking at only the two non-aces and the four aces, with $N$ denoting non-aces, and $A$ denoting aces.

$\boxed{N}\boxed{N}\quad\boxed{A}\quad\quad\boxed{A}\quad\quad\quad\boxed{A}\quad\boxed{A}\quad Pr = \dfrac{48}{52}\dfrac{47}{51}\dfrac{10}{50}\dfrac9{49}\dfrac{8}{48}\dfrac7{47}$

The idea is that in sequence drawn, the first two must be non-aces, and there must be four aces somewhere in the next ten.

Further Simplification

Noting @JMP's comment, a further simplification is possible.

The "strip" of $10$ containing $4$ aces and $6$ non-aces could start anywhere and would necessarily contain two consecutive non-aces on at least one end, thus we can simply write

$Pr = \dfrac{10}{52}\dfrac{9}{51}\dfrac{8}{50}\dfrac{7}{49}$

Your answer is correct !

Here is just another way looking at only the two non-aces and the four aces, with $N$ denoting non-aces, and $A$ denoting aces.

$\boxed{N}\boxed{N}\boxed{A}\boxed{A}\boxed{A}\boxed{A}\quad Pr = \dfrac{48}{52}\dfrac{47}{51}\dfrac{10}{50}\dfrac9{49}\dfrac{8}{48}\dfrac7{47}$

The idea is that in sequence drawn, the first two must be non-aces, and there must be four aces somewhere in the next ten.

Your answer is correct !

Here is just another way looking at only the two non-aces and the four aces, with $N$ denoting non-aces, and $A$ denoting aces.

$\boxed{N}\boxed{N}\quad\boxed{A}\quad\quad\boxed{A}\quad\quad\quad\boxed{A}\quad\boxed{A}\quad Pr = \dfrac{48}{52}\dfrac{47}{51}\dfrac{10}{50}\dfrac9{49}\dfrac{8}{48}\dfrac7{47}$

The idea is that in sequence drawn, the first two must be non-aces, and there must be four aces somewhere in the next ten.

Your answer is correct !

Here is just another way looking at only the two non-aces and the four aces, with $N$ denoting non-aces, and $A$ denoting aces.

$\boxed{N}\boxed{N}\boxed{A}\boxed{A}\boxed{A}\boxed{A}\quad Pr = \dfrac{48}{52}\dfrac{47}{51}\dfrac{10}{50}\dfrac9{49}\dfrac{8}{48}\dfrac7{47}$

The idea is that in sequence drawn, the first two must be non-aces, there must be four aces somewhere in the next ten, and what fills the rest is immaterial.

Your answer is correct !

Here is just another way looking at only the two non-aces and the four aces, with $N$ denoting non-aces, and $A$ denoting aces.

$\boxed{N}\boxed{N}\boxed{A}\boxed{A}\boxed{A}\boxed{A}\quad Pr = \dfrac{48}{52}\dfrac{47}{51}\dfrac{10}{50}\dfrac9{49}\dfrac{8}{48}\dfrac7{47}$

The idea is that in sequence drawn, the first two must be non-aces, and there must be four aces somewhere in the next ten.