Probability of getting >70% in exam with 50 yes/no questions In a paper containing 50 yes/no questions, I am trying to find the probability of getting 70%.
Using binomial distribution,
$$P(X\ge70\%)=\sum_{k=25}^{50} \binom{50}{k}\left(\frac{1}{2}\right)^{50}$$
The following result was obtained:
70% (Grade A in university) in Assessment: 0.199913% chance
I am not sure I have followed the formula correctly so looking for approval and guidance.
 A: The expression
$$\sum_{k=25}^{50} \binom{50}{k}\left(\frac{1}{2}\right)^{50}$$
gives the probability of getting $25$ or more correct answers on a $50$-question True/False test, if one tosses a fair coin on each question to choose the answer.
It can be evaluated exactly using various tools. The sum is approximately $0.556$. 
For this particular sum, there is a useful shortcut. By symmetry, the sum $a$ from $0$ to $24$ is the same as the sum from $26$ to $50$. It follows that 
$$2a+\binom{50}{25}\left(\frac{1}{2}\right)^{50}=1.$$ 
Thus our sum is equal to $\dfrac{1}{2}+ \dfrac{1}{2}\dbinom{50}{25}\left(\dfrac{1}{2}\right)^{50}$.
Remark: For the probability of a grade of $70\%$ or higher, we would add up from $35$ to $50$, not $25$ to $50$. The result is about $0.0033$.  
A: The binomial distribution is given by
$$P(X = n) = {50 \choose n} \frac{1}{2^{50}}.$$    
The probability of obtaining more successes than the $N$ observed in a binomial distribution is 
$$ \sum_{n = N}^{50} P(X = n). $$
So if you want to get the probability to get grade A randomly, you should compute
$$ \sum_{n = 35}^{50} {50 \choose n} \frac{1}{2^{50}} $$
and its answer is approximately $0.3\%$ according to WolframAlpha.
It is hard to compute such summation without computer. If you want to do it by hand, normal approximation may helpful.
