Probability of of an event happening at least once in a sequence of independent events? I want to find the probability of flipping heads at least once if you flip a coin two times. The possible outcomes (we don't care about the order) are (each equally likely)  TT, TH, HT, HH. Three out of four have an H in them, so the probability is $\large \frac 34$. Is this correct? Is there a better and efficient way (especially when dealing with a higher number of flips? Please use only very basic terminology and concepts from probability because I've never taken a class. Thanks.

We have the following equally likely outcomes:
T T 
T H <-- 
H T <--
H H <--
In three of the four outcomes, a Heads appears: Probability of at least one head is indeed $\dfrac 34$. Put differently, note that the probability that no heads appears is 1 out of four or $\frac 14$. So the probability of at least one head is $1$ minus the probability of getting NO heads, which is $1$ minus the probability of getting all tails: is $1 - \frac 14 = 1 - \frac{1}{2^2} = 1 - \frac 12 \cdot \frac 12$
In the above demonstration, it is quite easy to list out the "probability space": which is essentially, all possible outcomes.
The "at least one head" qualifier is handy, because it allows you to simplify the determination of probability for any arbitrarily large numbers of flips.
If you flip a coin $n$ times, and want to know the probability of getting at least one head, note the outcome of getting all tails is the "complement" of the set of outcomes in which you get at least one head. The probability of getting all/only tails, when flipping a coin $n$ times is equal to $$\underbrace{\dfrac 12 \cdot \frac 12 \cdot \cdots \cdot \frac 12}_{\large n\; \text{times}} = \left(\frac 1{2}\right)^n$$
So...$$\begin{align} \text{probability of getting at least one head}\; & = 1 -\text{probability of not getting any heads}\; \\ \\ & =  1 - \;\text{probability of getting all tails}\;\\ \\& = 1 - \left(\dfrac{1}{2}\right)^n\end{align}$$
