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.
The number $X$ of correct answers in the remaining $75$ that were answered randomly, is binomially distributed with parameters $n=75$ and $p=0.25$, in symbols $$X \sim B(n=75, p=0.25)$$ The probability to pass the exam is equal to $$P(X\ge23)=\sum_{x=23}^{75}P(X=x)$$ Since it is difficult to calculate the above sum if you are not allowed to use a calculator or a program like excel, you can approximate $X$ with a normal random variable $Y$ such that $$Y \sim N(μ=np, σ^2=np(1-p)) \implies Y \sim N(μ=18.75, σ^2=14.065)$$ Now $$P(X\ge 23)=1-P(Y\le 22.5)$$ where we applied also the continuity correction.
Best Answer
I would say, will determine the use of the central limit theorem (using tables of the normal probability distribution - approximation to the binomial distribution Bin(n,p) = Bin(40,0.2) normal distribution $N(\mu,\sigma)=N(n*p,\sqrt{np(1-p)}=N(40\cdot 0.2,\sqrt{40 \cdot 0.2 (1-0.2)}$).