What a quite bizarre concept. Are there different probabilites of a ball landing in any particular compartment? I'm guessing there is, since if there were an equal change, all 8 balls landing in Compartment 1 (scoring 8) would have exactly the same likelihood of occuring as all 8 balls landing in Compartment 2 (scoring 16), but the second apparently doesn't win you any prizes.
I'm also going to assume that each ball is independent: if Compartment Y is filled with X balls, then ball X+1 has exactly the same chance of landing in Compartment Y as any preceding ball.
I've been thinking a lot more about your problem (even diving back into my combinatorics of several years ago), but I can't think how best to proceed, and I also don't know how your work needs to be set out. If I were doing it casually, for myself, not intending to show my workings out to anyone, I'd do it something like this:
P(1), P(7) = 0.05
P(2), P(6) = 0.1
P(3), P(5) = 0.2
P(4) = 0.3
P(1×7) = 0.057
P(1×6 + 2) = 0.056 × 0.1
P(1×5 + 3) = 0.055 × 0.2
And so on. I'm sure you can see what I'm doing. Trouble is it requires a hell of a lot of pencil mileage. Almost certainly this can be solved with combinations (C75) or permutations (P75) somehow, but I can't remember how to use them properly, and if you haven't been introduced to them, your teacher wouldn't expect you to use them.
You could shorten this section by simply stating the general formula, which would be (given the probabilities I've invented here):
P(1a + 2b + 3c + 4d + 5e + 6f + 7g) = 0.05a+g × 0.1b+f × 0.2c+e × 0.3d
Where a to e are the numbers of balls in each compartment
You can then add together the probabilities of all the outcomes that yeild the same score, for example, 10:
P(Score=10) = P(1×6 + 4) + P(1×5 + 2 + 3) + P(1×4 + 2×3)
= (0.056 × 0.3) + (0.055 × 0.1 × 0.2) + (0.054 × 0.13)
= 1.71875×10−8
= 1.72×10−8 (3 s.f.)
Which seems a stupidly small amount, but that's because the probabilities I've used are greatly exaggerated over those a real-life device of this nature would exhibit.
Of course, if there is equal probability of any ball landing in any compartment, then P(Score=Z) = 1/7 × the number of ball arrangements that score Z
I hope this helps at least a little, it's been thrown together haphazardly.
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