Let's suppose there is a test of $N$ questions, each one with multiple choices answer, let's say $p$. To say if better, for every question there are $p$ possible answers, but one and only one is the correct one.
What is the probability to get a good score by answering them in a random way? (good score = at least $70-80$ %).
I also would like to generalise it, by introducing $N_i$ questions, in which each of them has a different number of multiple choice answer $p_i$, in which
Case 1: one and only one is the correct one;
Case 2: there could be one or 2 correct answers (not always $2$).
To be clear: $p_i$ represents the number of given answers, not the number of possible correct answers.
How could I proceed?
I thought that for the first question, in which we have $N$ questions with $p$ possible answers each one, we can safely state that there is one and only one way in which we can get $100$ %. So I have $\frac{1}{p}$ probability to answer correctly to the first question, $\frac{1}{p}$ for the second and so on.
The moral is: I have $$\frac{1}{p^N}$$
probability of answering good to all the question. That seems legit. But what if I ask for a total score of $80$ % ?
About the second case, I guess the same reasoning holds, namely: $\frac{1}{p_1}$, $\frac{1}{p_2}$ and so on.. hence:
$$P = \prod_{k = 1}^N\frac{1}{p_i}$$
And if I have the possibility of having two correct answers, I shall introduce a sort of bias $b_i$ such that
$$b_i = \begin{cases} 1 \\ 2 \end{cases}$$
Depending on the question. Example: if the second question has $4$ possible answers but only $2$ of them are correct, then I'll have $1/2$ probability in guessing them correctly.
This suggests me the probability for each question is
$$P_i = \frac{b_i}{p_i}$$
So the total probability should be:
$$P = \prod_{i = 1}^N \left(\frac{b_i}{p_i}\right)$$
Does that hold? Or it's a load of malarkey?
And how could I manage to get the desired result of $80$ %?