Probability of hitting a rectangular target given deviation of the bullet

Question

The target is a $3 \times 2$ rectangle. A shooter aims at a random point in the target. Deviation of the bullet from that point is random but does not exceed 0.1 on each axis (assume the sides of the target are parallel to the axes). What is the probability that the shooter hits the target?

My approach I split the target into central part and (thick) edges. Probability of hitting the central part is 1. Edges are sides and angles. Probability of hitting the target on a side is from 1 to 1/2, and on an angle is from 1 to 1/4. Now just taking the integrals and weighing them by respective probabilities of aiming at those areas should give the answer.

(1) Is this approach correct?

(2) Is there a more elegant way to deal with this problem?

Answer 1

EDIT: This is actually not right, see the comments below!

The part of "deviation of the bullet from that point is random but does not exceed 0.1 on each axis" does not confirm what shape the bullet can deviate to from the targetted point on the rectangle. This can either be a circle or a square. Ill calculate both of the versions and I'll also assume that the deviation is Unoform on the given Circle or Square.

Version 1:

$$\Bbb{P}(the\ bullet\ hits) \stackrel{*}{=} \frac{area\ of\ starting\ rectangle}{all\ possible\ area\ to\ hit} =$$ $$= \frac{3 \cdot 2}{3 \cdot 2 + \underbrace{2(2 \cdot 0.1)}_{green\ area}+\underbrace{2(3 \cdot 0.1)}_{blue\ area} + \underbrace{4\frac{0.05^2\pi}{4}}_{4\ quarter\ circles}} = \frac{6}{6+0.4+0.6+ \sim 0.007854} \approx$$ $$ \approx 0.8562 = 85.62\%$$

$*$ (assuming that the deviation is Uniform)

Version 2:

$$\Bbb{P}(the\ bullet\ hits) \stackrel{*}{=} \frac{area\ of\ starting\ rectangle}{all\ possible\ area\ to\ hit} =$$ $$= \frac{3 \cdot 2}{3.2 \cdot 2.2} = \frac{6}{7.04} = \frac{75}{88} \approx 0.8523 = 85.23\%$$

Answer: Assuming the deviation is circular: $85.62\%$, assuming the deviation is a square: $85.23\%$.