Probabilities of at least one head with a (very) unfair coin 
I have a large pool of DNA molecules. A few carry a mutation, but most
do not. For each DNA molecule,
$$P(\text{mutated}) = 0.0003$$
$$P(\text{normal}) = 0.9997$$
I am drawing $300$ of those DNA molecules. What is the probability
that I take at least one mutated?

My logic is that this is like flipping a (very) unfair coin $300$ times, where $P(\text{heads}) = 0.0003$ and $P(\text{tails}) = 0.997$. So what is the probability of getting at least one head (mutation)? This is equal to the probability of not getting all tails (normal).
There are only two possibilities: I get all tails (normal), or I do not. So we have the probability of getting at least one mutation is:
$= 1 - P(\text{all normal})$
$= 1 - P(\text{normal}) \cdot P(\text{normal}) \cdot P(\text{normal}) \cdots \text{($300$ times)}$
$= 1 - P(\text{normal}) ^ {300} $
$= 1 - 0.91$
$= 0.09$
Hence, there is a $9\%$ chance of getting at least one mutation.

Now, there are only a few options: I can have exactly $1$ mutation, or I can have exactly $2$ mutations, etc. My intuition tells me that summing the probabilities for all these options should give the same result, i.e., the probability of getting exactly $1$ mutation $+$ the probability of getting exactly $2$ mutations $+ \cdots$ should give $0.09$.
Probability of getting exactly $1$ mutated
$=P(\text{mutated}) \cdot P(\text{normal}) \cdot P(\text{normal}) \cdot \cdots \cdot P(\text{normal}) \quad \text{($299$ normal occurrences)}$
$= P(\text{mutated}) \cdot P(\text{normal})^{299}$
$= 0.0003 \cdot (0.9997)^{299}$
$= 0.00027$
So the logic seems to be:
$$\text{Probability of getting exactly $n$ mutated} = P(\text{mutated})^n * P(\text{normal})^{300-n}$$
With this formula, I can calculate the probability of getting exactly $2$ mutated, $3$ mutated, $4$ mutated, etc., until $300$. However, when I sum all these probabilities (done in R), I only get $0.000274$. Where did I go wrong?
 A: Great approach, but it has one error. You should take the positions of mutations into account. The value:
$$P(\text{mutated}) \cdot P(\text{normal})^{299}$$
is equal to the probability of having one mutation at a specific position. For instance, this is the probability of having one mutation , which is at the first position.
You must consider the number of positions where the mutation might appear to fix your formula. Here, it is obvious that there are $300$ possible positions for the mutation. In general, if you have $n$ mutations, the total number of ways to choose $n$ positions from a total of $m$ positions (here $m=300$)  is given by:
$${m \choose n} = \frac{m!}{n!(m-n)!}$$
These are known as binomial coefficients, and the above is read as '$m$ choose $n$'. You should read up on them!
The correct probability of $n$ mutations is hence given by:
$${300 \choose n}P(\text{mutated})^n \cdot P(\text{normal})^{300-n}$$
Here's a way you can make sense of this coefficient. Assume you have $$P(\text{mutated}) = P(\text{normal}) = 0.5$$ just like a coin toss. Your original formula would suggest that the probability of getting no heads and $150$ heads are equal, which is not true. We expect roughly half the tosses to land heads with high probability. This is because the coefficient ${300 \choose n}$ is highest when $n=150$.
