Number of questions to attempt to pass a multiple choice test 
There are $100$ multiple choice questions in a test, with $4$ options each.  The probability of choosing a right answer for a question is $0.25$.  Each right answer gets $4$ marks and each wrong answer gets $-1$ mark.  The passing criteria is $50\%$. How many questions should a person attempt to pass the test, given that they are sure of getting at least $30$ questions right.

I tried:
To get $50\%$ (200 marks) in the test,
\begin{align*}
(100 - x)4 -x & = 200\\
400 - 5x & = 200\\
5x & = 200\\
x & = 40 
\end{align*}
incorrect
Should the person attempt $60$ questions to pass the test?
 A: The formulation of the question is a little bit odd to me, since whatever the number of questions the person answers, there is a possibility of failing. Obviously, the person will get $120$ points from the questions they are sure of knowing the answer to, so they need $80$ additional points out of the $70$ other questions. They may answer $20$ questions and luckily get them all correct, or answer $70$ questions and get the all wrong, losing $70$ points. What we can do is to compute the probability of passing the test as a function of the number of questions answered randomly, and mazimize this probability
Assume that the person answers randomly $n$ questions, where $0\le n\le 70$ and $Y$ the number of correct answers to these questions. Then $Y$ has a binomial distribution of parameters $n$ and $1/4$. Their score will be
$$X=120 + 4Y + (-1)(n-Y) = 120-n + 5Y$$
so the probability $p(n)$ of passing the test is
$p(n)=P(X\ge 200) = P(120-n+5Y \ge 200) = P(Y \ge 16 + \frac{n}{5})= \sum_{\lceil 16+\frac{n}{5} \rceil}^n \binom{n}{k}\left(\frac{1}{4}\right)^k\left(\frac{1}{4}\right)^{n-k}$
Using a computer, we see that $p(n)$ is very small for every $n$ and it is maximum when $n=70$ with a whooping $p(70)=0.000824$ probability of passing the test. They'd better study than rely on luck to pass the exam.

We may also count the average number of points the person will make on the test by answering randomly $n$ questions. Using the notations above, we have $E(Y)=\frac{n}{4}$ so
$$E(X)=120-n+ 5E(Y) = 120 + \frac{n}{4}$$
We have $E(X)\ge 200 \iff n \ge 320$. This explains why the probability of passing the test is extremely low. Intuitively, each time the person answers $4$ questions randomly, they will lose $3$ points and gain $4$ so a net gain of only $1$ point.
