How to attack this probability question directly? Below is a example probability question I found in my book. I want to attack this question directly instead of using the complement approach. How do I do that? My book always uses the complement approach, but I want to see how to tackle it head on. 
Example Problem:
Mark is taking four final exams next week. His studying was erratic and all scores A, B, C, D, and F are equally likely for each exam. What is the probability that Mark will get at least one A?
Take complements. The complementary event of getting at least one A is getting no As. Since outcomes are equally likely, by the multiplication principle there are $4^4$ exam outcomes with no As (four grade choices for each of four exams). And there are $5^4$ possible outcomes in all. The desired probability is $1−(4^4/5^4)=0.5904$.
 A: The probability of a union of disjoint events is the sum of the probabilities of the events.
The event "$E$: getting at least one A" is the union of the disjoint events "$E_1$: getting exactly one A," "$E_2$: getting exactly 2 A's," "$E_3$: getting exactly 3 A's," and "$E_4$: getting exactly 4 A's."
Then $P[E]=P[E_1]+P[E_2]+P[E_3]+P[E_4]$.
Do you know how to get $P[E_1]$ and the other probabilities?
A: There's a reason taking complements is often suggested.  But to tackle it head on, as you say, we note that the number of A is a random variable with a binomial distribution of ${\cal Bin}(4, 1/5)$   (Do you know why?)
$$X_A\sim{\cal Bin}(n, p) \iff \mathsf P(X_A=k) = {n\choose k}p^k (1-p)^{n-k}$$
So for this problem, $n=4, p=1/5$ and we want $X_A\geqslant 1$ so we need to evaluate the series:
$$\begin{align}\mathsf P(X_A\geqslant 1) &= \sum_{k=1}^4 \mathsf P(X_A=k) \\ & = {4\choose 1}\frac {4^3}{5^4} + {4\choose 2}\frac {4^2}{5^4} + {4\choose 3}\frac {4}{5^4} + {4\choose 4}\frac {1}{5^4}
\\[4ex]
\mathsf P(X_A\geqslant 1) & = 1-\mathsf P(X=0)
\\ & = 1- {4\choose 0}\frac {4^4} {5^4}
\\ & = \frac{369}{625}
\end{align}$$
