Probability of passing a T/F exam? There are $n$ questions in an examination ($n \in \mathbb{N}$), and the answer to each question is True or False. You know that exactly $t$ of the answers are True ($0 \le t \le n$), so you randomly answer $t$ questions as True and the rest as False. What is your probability of getting at least a $50\%$ score in the exam (in terms of $n$ and $t$)?
My attempt:
WLOG assume you answer the first $t$ questions as True. Let there be $k$ questions out of the first $t$ of which the answer is True. Thus, out of the other $n-t$ questions where you replied False, $(n-t)-(t-k)=n-2t+k$ questions are really False. Thus, the fraction of correct answers for the whole examination is equal to $\frac{n-2t+2k}{n}$. If this is at least $1/2$, then $n+2k \ge 4t$. However, I can't calculate the probability of this happening.

There is a typo. Instead of $n+2k \ge 4t$, it should be $n+4k \ge 4t$. I do not know if you can get a closed form but here is my work that may help you understand lower and upper bound of $t$ in terms of $n$, beyond which one is certain to get at least $50\%$ score.
a) Based on your work, $ \displaystyle k \geq \frac{4t - n}{4}$. So if $t \geq \dfrac{3n}{4}, k \geq \dfrac{n}{2}$ and we are bound to get at least $50\%$ score by choosing $t$ answers as TRUE.
Similarly if $ ~\displaystyle  t \leq \frac{n}{4}$, you are certain to get at least $50\%$ score by randomly choosing $(n-t)$ answers as FALSE.
b) Now for $~ \displaystyle  \frac n4 \lt t \lt \frac {3n}4$,
we must have $ \displaystyle \lceil \frac{4t - n}{4} \rceil \leq k \leq t$ to score at least $50\%$
So the desired probability can be written as,
$ \displaystyle \sum \limits_{k = k_l}^t {t \choose k} {n - t \choose t - k} / {n \choose t}$
where $ \displaystyle k_l = \lceil \frac{4t - n}{4} \rceil$
