Set coordinates:

E=(0,0)
D=(9,0)
F=(0,12)

Then \(D F=\sqrt{9^2+12^2}=15\) .

Let T be a point on D F . Parametrize:

$$
T=(9-9 t, 12 t), \quad 0 \leq t \leq 1
$$

Then

$$
E T^2=(9-9 t)^2+(12 t)^2=225 t^2-162 t+81
$$

We need \(E T<9 \), so:

$$
\begin{gathered}
225 t^2-162 t+81<81 \\
225 t^2-162 t<0 \\
9 t(25 t-18)<0
\end{gathered}
$$

Since \(t \geq 0\) , this gives:

$$
0<t<\frac{18}{25}
$$


So the portion of the hypotenuse where $E T<9$ has length:

$$
\frac{18}{25} \cdot 15=10.8
$$

Total hypotenuse length is 15, so:

$$
\text { Probability }=\frac{10.8}{15}=0.72
$$

Quantity A is 0.72, which is greater than Quantity B 0.7.

A