Probability of passing test by choosing answers randomly Can anybody help us calculate the overall probability to pass a multiple choice test by guessing randomly (minimum score of 60% required) with 20 questions of type A (probability of 0.2 each) and 10 questions of K-Prim where I can't calculate the exact probability.
K-Prim-Questions have 4 answers where you can choose between TRUE and FALSE. If you get 3 out of 4 right, you're given 0.5 points, if you get all of them right, 1 point. 2 or beneath yield 0 points.
Each question (Type A and Kprim) yields 1 point max.
Overall the test gives you 30 points max.
We tried to use Bayes' Theorem but since the K-Type-probabilty is unknown, I can't proceed.
Any help is appreciated.
Cheers
 A: The idea is to construct a probability distribution on the 10 K-Prim type questions.  The probability distribution for a single question is $$\begin{align*} p_0 = \Pr[S = 0] &= \sum_{k=0}^2 \binom{4}{k} (0.5)^k (0.5)^{4-k} = \frac{11}{16}, \\ p_1 = \Pr[S = 0.5] &= \binom{4}{3} (0.5)^3 (0.5)^1 = \frac{1}{4}, \\ p_2 = \Pr[S = 1] &= \binom{4}{4} (0.5)^4 (0.5)^0 = \frac{1}{16}. \end{align*} $$  This assumes that for each such question, the choice of True/False is equally likely for each of the four answers, and the each answer is independent, thus the number of correct answers follows a ${\rm Binomial}(4,0.5)$ distribution.
Next, the distribution of the sum of the scores of 10 K-Prim questions can be derived from the multinomial distribution, though it is somewhat tedious to compute:  let $X_0$, $X_1$, $X_2$ be random variables that count the number of $0$-point, $0.5$-point, and $1$-point scores out of the 10 questions.  Then $$\Pr[(X_0, X_1, X_2) = (a,b,c)] = \frac{10!}{a! b! c!} p_0^{a} p_1^b p_2^c.$$  Then we can tabulate the sum; we do this in Mathematica:
Flatten[Table[{b/2 + c, PDF[MultinomialDistribution[10, {11/16, 1/4, 1/16}], 
        {10 - b - c, b, c}]}, {b, 0, 10}, {c, 0, 10 - b}], 1]

Table[{k, Total[Select[%, #[[1]] == k &]][[2]]}, {k, 0, 10, 1/2}]

which gives us the desired probability distribution for these 10 questions.  Call this random variable $K$.  Now, for the remaining 20 questions, the total point count is simple; it is simply $A \sim {\rm Binomial}(20, 0.2)$.  So the probability that the total score is at least $18$ out of $30$ is $$\sum_{k=0}^{20} \Pr[K = k/2]\Pr[A \ge 18 - k/2].$$  Again, we use Mathematica:
Sum[%[[k, 2]] (1 - CDF[BinomialDistribution[20, 1/5], 18 - k/2]),
    {k, 1, Length[%]}]

This gives us $$\frac{8327843221553613}{2^9 \cdot 10^{20}} \approx 1.62653 \times 10^{-7}.$$  This is so small that it is unlikely that a naive simulation approach will be able to approximate it.
A: Quick and dirty simulation...
For the 20 multiple choice question choose a right answer from the 5 available at random, the right answer is number 1.
fA20[] := Count[RandomInteger[{1, 5}, 20], 1]

For the K-Prim type, flip a coin for each of the four answers, add up the number that are correct and score that total according to the rules.
fKPrim10[] :=  Total[Table[   Total[RandomInteger[{0, 1}, {4}]] /. {4 -> 1, 3 -> 0.5, 2 | 1 -> 0}, {10}]]

Run this many times to get a distribution of scores...
Histogram[Table[fA20[] + fKPrim10[], {10000}]]

