19. Let circles $\Gamma_{1}$ and $\Gamma_{2}$ be two concentric circles with center $O$, and radii $2$ and $1$, respectively. Then the probability that a randomly chosen chord of the larger circle intersects the smaller circle is ( ).
(A) $\frac{1}{6}$
(B) $\frac{1}{4}$
(C) $\frac{2-\sqrt{2}}{2}$
(D) $\frac{1}{3}$
(E) $\frac{1}{2}$
19. D.

As shown in Figure 6, take any point $A$ on the circumference of the larger circle $\Gamma_{1}$. Draw two tangents $AB$ and $AC$ from $A$ to the smaller circle $\Gamma_{2}$, intersecting the circle $\Gamma_{1}$ at points $B$ and $C$, respectively, and connect $BC$. The chord $AA'$, which intersects the circle $\Gamma_{2}$, has its other endpoint $A'$ on the minor arc $\overparen{BC}$ (excluding points $B$ and $C$).

By symmetry, we know that $\triangle ABC$ is an isosceles triangle.

Since $AB$ and $AC$ are tangents to the circle $\Gamma_{2}$, the distances from point $O$ to the sides $AB$ and $AC$ are both 1.
Also, $AO=2$, so $\angle BAO = \angle CAO = 30^{\circ}$.
Therefore, $\triangle ABC$ is an equilateral triangle.
Thus, the minor arc $\overparen{BC}$ is $\frac{1}{3}$ of the circumference of the circle $\Gamma_{1}$.
Hence, the probability of the chord $AA'$ satisfying the condition is $1 \times \frac{1}{3} = \frac{1}{3}$.