The diagram shows a goat tied to a pole at A.
The length of the rope attached to the goat is 6 m.
A is due west of the centre of the circle, O.

1. Identify the two circles and find the angles needed for the intersecting sectors

Let the two circles intersect at $P$ and $Q$.

For the circle centred at $A$, radius $6$ m.
For the circle centred at $O$, radius $4.5$ m.
Also, $AO=4.5$ m.

In triangle $AOP$,
$AP=6,\quad OP=4.5,\quad AO=4.5.$

Using the cosine rule at $A$,
$\cos \angle OAP=\frac{6^2+4.5^2-4.5^2}{2(6)(4.5)}=\frac{2}{3}$
$\angle OAP=48.19^\circ.$

So the sector angle at $A$ is
$2(48.19^\circ)=96.38^\circ.$

Using the cosine rule at $O$,
$\cos \angle AOP=\frac{4.5^2+4.5^2-6^2}{2(4.5)(4.5)}=\frac{1}{9}$
$\angle AOP=83.62^\circ.$

So the sector angle at $O$ is
$2(83.62^\circ)=167.24^\circ.$

2. Find the overlapping area of the goat's reachable region and the grass patch

Area of overlap
$=\text{sector at }A+\text{sector at }O-\text{two triangles}.$

$\text{sector at }A=\frac{96.38}{360}\pi(6)^2$

$\text{sector at }O=\frac{167.24}{360}\pi(4.5)^2$

Area of triangle with sides $6,6$ and included angle $96.38^\circ$:
$\frac12(6)(6)\sin 96.38^\circ$

Area of triangle with sides $4.5,4.5$ and included angle $167.24^\circ$:
$\frac12(4.5)(4.5)\sin 167.24^\circ$

Hence
$\text{overlap area}$
$=\frac{96.38}{360}\pi(6)^2+\frac{167.24}{360}\pi(4.5)^2-\frac12(6)(6)\sin 96.38^\circ-\frac12(4.5)(4.5)\sin 167.24^\circ$
$=39.7\text{ m}^2\text{ approximately}.$

3. Divide by the total possible area the goat can reach

Total area the goat can reach:
$\pi(6)^2=36\pi.$

Therefore,
$\text{Probability}=\frac{39.7}{36\pi}=0.351\text{ approximately}.$