If there are 100 MCQs with 4 options each. The probability that a person gets an question right is 0.25. What is the probability of getting at least 50 questions of 100 right?
What is the probability of getting at least 50 questions of 100 right?
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The number $X$ of questions correct if the student is answering strictly at random (with no knowledge of the subject matter) has $X \sim \mathsf{Binom}(n = 100,\, p = 1/4).$
So the probability of getting exactly fifty questions correct is $$P(X = 50) = {100 \choose 50}\left(\frac 14\right)^{50}\left(\frac34\right)^{50} = 4.5073 \times 10^{-8}.$$ Computation using R statistical software:
dbinom(50, 100, 1/4)
## 4.507311e-08
The probability of getting at least fifty correct is
$$P(X \ge 50) = \sum_{k=50}^{100} {100 \choose k}\left(\frac 14\right)^{k}\left(\frac34\right)^{100-k} = 6.6385 \times 10^{-8}.$$
1 - pbinom(49, 100, 1/4)
## 6.638502e-08
Both probabilities are very small because most of the probability in the distribution $\mathsf{Binom}(n = 100,\, p = 1/4)$ is centered near $\mu = E(X) = np = 100(1/4) = 25.$
Here is a figure that shows the distribution of $\mathsf{Binom}(n = 100,\, p = 1/4)$ along with the density function of $\mathsf{Norm}(\mu = 25, \sigma = 4.33),$ where $\sigma = \sqrt{np(1-p)} =$ $\sqrt{75/4} = 4.3301.$
Unless you are using software (or a statistical calculator) in your class, my guess is that you are supposed to use the normal approximation to the binomial distribution to approximate the very small value of $P(X \ge 50).$ (As @LordShark commented.) I will show the start of that procedure, and let you verify it and finish it for yourself:
$$P(X \ge 50) = P(X > 49.5) = P\left(\frac{X - \mu}{\sigma} > \frac{49.5 - 25}{4.3301} \right)\\ \approx P(Z > 5.658) = ?$$ where $Z$ is a standard normal random variable.

Hint: The number of ways to get $n$ questions right is $\binom{100}{n}$, and given any set of $n$ questions to get right the probability of doing so is
$$(0.25)^n(0.75)^{100-n}.$$
Can you solve it from there?