I saw no answer explicitly using the surface of a sphere for an argument, so here goes:
Hitting a target can be thought of as hitting a shape the size of the target on a sphere with you in the center and a radius of target distance. The probability of hitting said shape while shooting wildly into any direction then is the ratio of target-area vs sphere-surface. Sphere surface is 4*pi*r², but i'll round 4*pi to 10 because it's not going to make much of a difference.
Thus, at 10m distance (in space. no gravity, and you can shoot any direction you want) the probability of hitting a 1m² target (~human or thermal vent of the Death Star) by shooting blindly is 1/1000 ( = 1m²/(10*(10m)² )- not bad. Because of the r² in the denominator, this probability will drop sharply.
At 1km (10^3 m) the probability is at 1/10 000 000, which is about jackpot-level in a small national lottery. (Here a note about statistics: If you take 10 million independent shots, this does not mean you are guaranteed to hit the target- i you fill out 10 million lottery tickets, you'll make sure that you pick different numbers, independent shots are comparable to buying 10 million lottery tickets and then letting monkeys fill them, you'll get lot's of doubles; -- To compute the probability of hitting a human in space at 1km distance by 10 million independent, random, shots, you have to take the probability of not hitting him with one shot nh = (9 999 999/10 000 000), multiply this with itself by the number of shots snh = (nh^10M) and substract from one h = (1- snh); You then have the probability of hitting that poor guy once or more. That probability is about 2/3. Again, not bad, but you are now shooting 10 million bullets....
At 1AE (distance earth - sun, harshly rounded to 10^12 m) the one-shot probability is 1/10^25, or much more impressively 1/10000000000000000000000000, if you say the target is 1000m² big (instead of 1m²) you are allowed to take three zeros off that (1/10^22). Probability of hitting this station-size target with 10M random shots is 1/10^15 (or about as big as winning a mid-size national lottery twice, in a row, with only one ticket per draw). --- Hitting a specific moon (~ earth moon, 1^12m² target surface) gets your one-shot probability up to 1/10^13, and you 10M-shot probability up to 1/1M ! (Though if a bullet hits a moon, and no one notices... did it hit?)
'Being on the outskirts of the solar system' might mean 10AE distance, bringing the one-shot-station-hit-probability down by another 100 to 1/10^24
Taking gravity into account changes the numbers only slightly (in a civilizatory timeframe, i.e excluding multiple passes of objects on huge orbits)- Shooting away from the sun at less then escape velocity will bend the bullets back, thereby about doubling the bullet density sun-wards. So instead of 1/too_much, you now have 2/too_much.
Lasers are out because the beam gets too wide to damage stuff, and nuclear rockets are not that much different from bullets; Assuming they explode near the unlucky non-target (if not aimed that way why should they, though?), their kill-radius is now the effective target-surface, otherwise the equation is just like with the bullets, but you'll probably not spray 10M nuclear warheads around in an encounter.
One last number fest: Assume there are 10 battles a year, with 10 combatants, each spraying 10M bullets wildly, for 10 years; Further assume there are 1000 stations with 1000m2 target area each, floating 10AE away (never occluding each other for ease of calculation). --- 10^12 bullets at a combined target surface of 10^6 m² in a sphere of 10^27 m² : 1-(( ((1- ((10^6)/(10^27)) ))^(10^12) ) = 1/10^9 - actually not as bad as i'd assumed coming in !
Please heed the note about probabilities given above! While winning the powerball lottery has a 1/10^8 chance, and there have been winners, this is because a lot of people play, and have been for some time. Number of 'players' (stations) and time have already been factored into many of my above values, so they are not directly comparable (Or rather, comparing them you have to keep in mind the givens: the last of my numbers, 1/10^9, is the probability of any station being hit at any time, by any of the bullets shot during 10 years, while the powerball probability of 1/10^8 is the probability of a specific player winning with a specific ticket at a specific draw...)