Points $A$, $B$, $C$, and $D$ are located on $\overline{AB}$ such that $AB = 3AD = 6BC$. If a point is selected at random on $\overline{AB}$, what is the probability that it is between $C$ and $D$? Express your answer as a common fraction.

[asy]

draw((0,0)--(12,.0000000001));
dot((0,0)); dot((4,0)); dot((10,0)); dot((12,0));
label("$A$", (0,0), S);
label("$D$", (4,0), S);
label("$C$", (10,0), S);
label("$B$", (12,0), S);

[/asy]
Since $D$ and $C$ are located on segment $\overline{AB}$, if $AB=3AD$, then $\overline{AD}$ must take up $1/3$ of line segment $\overline{AB}$. Similarly, since $AB=6BC$, $\overline{BC}$ must take up $1/6$ of line segment $\overline{AB}$. Then, $\overline{CD}$ is the remaining segment of $\overline{AB}$ and takes up $1-1/3 - 1/6 = 1/2$ of the total length of $\overline{AB}$. Thus, if we were to choose a random point on segment $\overline{AB}$, there would be a $\boxed{\frac{1}{2}}$ probability that it is between points $C$ and $D$.
\frac{1}{2}
Level 2
Counting & Probability
Problem: Points $A$, $B$, $C$, and $D$ are located on $\overline{AB}$ such that $AB = 3AD = 6BC$. If a point is selected at random on $\overline{AB}$, what is the probability that it is between $C$ and $D$? Express your answer as a common fraction.

[asy]

draw((0,0)--(12,.0000000001));
dot((0,0)); dot((4,0)); dot((10,0)); dot((12,0));
label("$A$", (0,0), S);
label("$D$", (4,0), S);
label("$C$", (10,0), S);
label("$B$", (12,0), S);

[/asy]
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the probability that a randomly selected point on line segment $\overline{AB}$ lies between points $C$ and $D$, we first need to determine the positions of points $C$ and $D$ based on the given relationships.

Let the length of segment $AB$ be denoted as $x$. According to the problem, we have the relationships:
1. $AB = 3AD$
2. $AB = 6BC$

From the first relationship, we can express $AD$ in terms of $x$:
AD = x/3.

From the second relationship, we can express $BC$ in terms of $x$:
BC = x/6.

Next, we can find the positions of points $D$ and $C$ on segment $\overline{AB}$:
- Since $A$ is at position 0 and $D$ is at position $AD$, we have:
D = 0 + AD = 0 + x/3 = x/3.

- Since $B$ is at position $x$ and $C$ is at position $BC$, we have:
C = B - BC = x - x/6 = 5x/6.

Now we need to find the length of the segment between points $C$ and $D$:
Length of segment $CD = C - D = (5x/6) - (x/3)$.

To perform the subtraction, we convert $x/3$ to a fraction with a common denominator of 6:
x/3 = 2x/6.

Thus, we have:
Length of segment $CD = (5x/6) - (2x/6) = 3x/6 = x/2.

Next, we calculate the total length of segment $\overline{AB}$, which is $x$.

The probability that a randomly selected point on $\overline{AB}$ lies between points $C$ and $D$ is given by the ratio of the length of segment $CD$ to the length of segment $AB$:
Probability = Length of segment $CD / Length of segment AB = (x/2) / x = 1/2.

Thus, the final answer is $\frac{1}{2}$.<|END_OF_LATENT|> The final answer is $\frac{1}{2}$.