Probability of passing a multiple choice "test" with multiple correct answers per question. I have created a "luck" game in the form of a multiple-choice quiz with 8 questions. 
Each question has 4 possible nonsense answers. The points awarded for each "correct" answer vary between 1-3. 
5/8 questions have more than 1 "correct" answer, although only one answer can be selected per question.
The possible points earned from each question are as follows:
A. 3/1/0/0 (where one answer is worth 3 points, one is worth 1 point, and two are worthless.)
B. 3/1/0/0
C. 2/1/0/0
D. 2/1/0/0
E. 2/1/0/0
F. 3/0/0/0
G. 3/0/0/0
H. 2/0/0/0
The highest achievable score is 20.
What is the probability of scoring 15 or higher, if answers are chosen at random?
 A: Since the score must be more than 14, we have to lose 5 points or less. As you can see for example in question A, we can either loose 2 marks, or 3 marks.(If we choose the option with 1 point, we have lost 2 points because the maximum mark possible is 3). So:A: -2   -3
B: -2   -3
C: -1   -2
D: -1   -2
E: -1   -2
F: -3
G: -3
H: -3
We have to lose either 5 marks, 4 marks, 3 marks, 2 marks, 1 mark or loose no mark at all.
5 = 2 + 3 = 1 + 2 + 2 = 1 + 1 + 3 = 1 + 1 + 1 + 2
4 = 1 + 1 + 2 = 2 + 2 = 1 + 3
3 = 1 + 1 + 1 = 1 + 2 = 3
2 = 1 + 1 = 2
1 = 1
0 = 0
Now the question can be easily solved. For example, if we want to loose exactly 5 marks, we have to loose 1 question with 2 marks and 1 question with 3 marks, or two questions with 2 marks and one question with one mark, or 2 questions with 1 mark and 1 question with 3 marks. In the former state, We have 6 questions with the possibility of loosing two marks and 3 with the possibility of loosing 3 marks.(Of course we have to notice that there are questions with both of the possibilities.) So the overall number of states which we loose 5 marks in them is:
5: ($\binom{3}{1}\binom{2}{1} + \binom{2}{1}\binom{1}{1}) + (\binom{3}{1}\binom{5}{1}) + (\binom{3}{2}\binom{3}{1}) + (\binom{3}{3}\binom{3}{1})$
We will do the same for the other 4 numbers. Then add up these numbers.
Now we have to divide this number by the number of all states which is obviously $4^8$.
A: Using computer simulation to simulate all possible combinations of answers to all 8 questions (there are 65,536 total possible outcomes), I am seeing 390 of them that total up to 15 or more points.  So the probability is $390/65536 = 0.00595$ which is about $6/10th$ of $1$%.  Not very good chances.
Just F.Y.I., I am including here the # of combinations for each possible score from 0 to 20:
$~~0 : ~~864$
$~~1 : 2160$
$~~2 : 3744$
$~~3 : 5832$
$~~4 : 7182$
$~~5 : 8247$
$~~6 : 8397$
$~~7 : 7668$
$~~8 : 6662$
$~~9 : 5132$
$10 : 3776$
$11 : 2525$
$12 : 1559$
$13 : ~~919$
$14 : ~~469$
$15 : ~~234$
$16 : ~~100$
$17 : ~~~~38$
$18 : ~~~~14$
$19 : ~~~~~~3$
$20 : ~~~~~~1$
The assumptions here are that all 8 questions must be answered but only 1 answer for each question can be chosen.
The simulation simply uses 8 nested loops which each go from 1 to 4 and those index 8 arrays which hold the points for each question-answer combo.  The program took me like 5 minutes to write and runs in about 1/6th of a second.
Notice that the most likely score would be 6 (out of 20) and also notice that no 2 different scores are equally likely (although 2 and 10 are very close).
