We are tasked with finding the probability that the distance from a randomly chosen point $P$ in the area of intersection of two circles is less than 3 units from the origin. The two given circles are:

• Circle $C_1$, centered at $(0, 0)$ with radius 5 units.
• Circle $C_2$, centered at $(4, 0)$ with radius 5 units.

Step 1: Equation of the circles

The equation of circle $C_1$ is: $x^2 + y^2 = 25.$ The equation of circle $C_2$ is: $(x - 4)^2 + y^2 = 25.$ We need to find the area of overlap between these two circles and determine the probability that a randomly chosen point in this region is within 3 units of the origin.

Step 2: Finding the area of intersection of the two circles

The two circles have the same radius (5 units), and the distance between their centers is 4 units. Since the sum of their radii is 10 units, which is greater than the distance between their centers (4 units), the circles intersect in a lens-shaped region.

We can use the formula for the area of intersection of two circles of equal radius $r$ with centers a distance $d$ apart: $A_{\text{intersection}} = 2r^2 \cos^{-1}\left(\frac{d}{2r}\right) - \frac{d}{2} \sqrt{4r^2 - d^2}.$ For our case, $r = 5$ and $d = 4$, so: $A_{\text{intersection}} = 2(5^2) \cos^{-1}\left(\frac{4}{2 \times 5}\right) - \frac{4}{2} \sqrt{4(5^2) - 4^2}.$ This simplifies to: $A_{\text{intersection}} = 50 \cos^{-1}\left(0.4\right) - 2 \sqrt{100 - 16}.$ $A_{\text{intersection}} = 50 \cos^{-1}(0.4) - 2 \sqrt{84}.$ Using a calculator, $\cos^{-1}(0.4) \approx 1.159$ and $\sqrt{84} \approx 9.165$, so: $A_{\text{intersection}} = 50 \times 1.159 - 2 \times 9.165 \approx 57.95 - 18.33 = 39.62 \, \text{square units}.$

Step 3: Finding the area where the distance from the origin is less than 3 units

Now, we need to find the area where the distance from the origin is less than 3 units, which is the area of the circle with radius 3 centered at the origin. The area of this smaller circle is: $A_{\text{small circle}} = \pi \times 3^2 = 9\pi \approx 28.27 \, \text{square units}.$

However, not all of this smaller circle lies within the intersection of the two circles. We need to find the region within the intersection where the distance from the origin is less than 3 units.

Step 4: Estimating the probability

The probability that a randomly chosen point in the intersection lies within 3 units of the origin is the ratio of the area of the smaller circle that lies within the intersection to the total area of the intersection. To estimate this, we need to approximate the area of the smaller circle within the intersection.

Since the smaller circle is centered at the origin and the distance from the origin to the center of $C_2$ is 4 units, the smaller circle will be partially outside the intersection. From geometric consideration, we can estimate that roughly half of the smaller circle lies within the intersection of the two circles.

Thus, the area of the smaller circle within the intersection is approximately: $\frac{1}{2} \times 9\pi \approx \frac{1}{2} \times 28.27 = 14.14 \, \text{square units}.$

Step 5: Calculating the probability

The probability is then the ratio of the area where the distance from the origin is less than 3 units to the area of intersection: $\text{Probability} = \frac{14.14}{39.62} \approx 0.357.$

Thus, the probability that the distance from the point to the origin is less than 3 units is approximately $\boxed{0.357}$.