Binominal distribution OR I got this exercise:

In a multiple choice quiz there are 5 questions and 4 choices for each question (a, b, c, d). Robin has not studied for the quiz at all, and decides to randomly guess the answers. What is the probability that:
she gets exactly 3 or exactly 4 questions right?

Should I understand this question as getting the EXACT value of 3 and EXACT value of 4, and then plus them together? Or is the question just calculating the EXACT for both 3 and 4? Right now I have just done a normal binomial calculation for exactly 3.
 A: The number of questions correct is $X \sim \mathsf{Binom}(n=5, p = 1/4).$ The four-place PDF (or PMF) table for this distribution can be found using R statistical software as follows (ignore row numbers in brackets [ ]):
x = 0:5;  pdf = round(dbinom(x, 5, 1/4), 4)
cbind(x, pdf)

     x    pdf
[1,] 0 0.2373
[2,] 1 0.3955
[3,] 2 0.2637
[4,] 3 0.0879  # P(X=3)
[5,] 4 0.0146  # P(X=4)
[6,] 5 0.0010

My interpretation of the English is that two separate
probabilities are requested. If asking for the sum, 
the question might have said "... either 3 or 4 questions right." But unless this is an online
computer quiz, there seems to harm in giving the two
numbers and their sum to be sure.
You should make sure you know how to use the binomial
PDF formula $P(X = k) = {n \choose k}p^k(1-p)^{n-k},$ for $k = 0, \dots, n.$ (I would not use a normal approximation for a binomial probability with such a small $n).$

A: You should understand the key conjunctions for the operations: AND means multiply, OR means add.
Thus, if we let $X$ be a random variable counting the number of correct answers, then $$P(X = 3 \ \text{or}\ X = 4) = P(X=3) + P(X=4).$$
