binomial-coefficientsprobabilitystatistics
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.
You can use binomial distribution that gives you the probability of $k$ successes out of $n$ trials when the success probability is $p$.
In your case, $n=50$ and $k$ varies based on the percentage of correct answers. For example, $100\%$ means $k=50$, i.e. $50$ correct answers with probability $p=0.5$(since two choices yes or no are made with equal probability). Can you solve now for rest of the grades?
Let $X$ be the random number of correct answers out of $n = 60$ true/false questions. The probability a single question is answered correctly is $p = 0.5$. Thus the exact distribution of $X$ is binomial: $$ X \sim {\rm Binomial}(n = 60, p = 0.5).$$ In order to get a passing grade of at least $70\%$, the required number of correct responses is $X \ge (0.7)(60) = 42$. Therefore, the desired probability is $$\Pr[X \ge 42] = \sum_{k=42}^{60} \binom{60}{k} (0.5)^k (1 - 0.5)^{60-k}.$$ This sum is tedious to compute (there are $60 - 42 + 1 = 19$ terms), so we use the normal approximation to the binomial distribution with continuity correction: if $np \ge 10$ and $n(1-p) \ge 10$ (which are satisfied in this case), then $$X \overset{\bullet}{\sim} {\rm Normal}\left(\mu = np, \sigma = \sqrt{np(1-p)}\right),$$ approximately. Hence to calculate $\Pr[X \ge 42]$, we compare the exact distribution of $X$ to a normal distribution with mean $\mu = 30$ and standard deviation $\sigma = 3.87298$. With continuity correction, in order to account for the entire discrete probability mass at $X = 42$, we must write $$\Pr[X \ge 42] \approx \Pr[X > 41.5] = \Pr\left[\frac{X - \mu}{\sigma} > \frac{41.5 - 30}{3.87298}\right] = \Pr[Z > 2.96929],$$ where $Z$ is a standard normal random variable. This results in a probability of $0.00149246$, very small. The exact value using the binomial distribution is $$\tfrac{1539401708565319}{1152921504606846976} \approx 0.00133522.$$ This probability is small because the standard deviation is small: for so many questions, it is extremely unlikely that one could answer much more than half correctly by random chance.
If you want to do these computations in a TI calculator, then the normal approximation is given by the command normalcdf(41.5,1E99,30,3.87298). The exact binomial probability is given by the command 1-binomcdf(60,0.5,41) because the complement of the event that $X$ is at least $42$ is that $X$ is at most $41$.
Best Answer
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 $0.0033$.