Q:

What's an intuitive or visual solution to calculate the probability to draw at least $k$ red balls after $n$ tries with replacement?

There is a bowl with red and blue balls.
The probability to draw a red ball is $p$.
Afterwards, the ball will be replaced, so the probability to draw a red ball won't change.
What's the probability to draw at least $k$ red balls after $n$ tries?

A:

From the question, immediately you notice ${k\leq n}$. It's an obvious fact, but worth stating (we are only picking one ball per try, so clearly the highest number of red balls we can get, $k$, is clearly less than or equal to the number of tries we perform, $n$).
We want to know the probability of getting at least $k$ red balls. This means you must take into account getting ${k,k+1,...,n}$ balls, also. For example, if ${n=10}$ and ${k=5}$, then if I get $6$ red balls after ${10}$ tries, then that is indeed at least $5$, and so must be included as part of our calculation.
So what we could do, is calculate the probability of getting exactly $k$ red balls, then calculate the probability of getting exactly ${k+1}$, then getting ${k+2}$... then getting ${n}$. What do we do with these probabilities? Well, we sum them up. So
$${P\left(\text{getting at least $k$ red balls after $n$ tries}\right)=\sum_{i=k}^{n}P\left(\text{probability of getting exactly $i$ red balls}\right)}$$
$${=P(\text{getting exactly $k$ red balls}) + P(\text{getting exactly ${k+1}$ red balls}) + ... + P(\text{getting exactly $n$ red balls})}$$
Now the only thing left to calculate is the probability of getting exactly ${k}$ balls (if we have this formula, obviously you can replace $k$ with ${k+1,k+2...}$ whatever you want).
Well, if I got ${k}$ red balls consecutively, then in the rest of the tries got blue balls (that is, in the first $k$ tries, I get red balls each time, then in the last ${n-k}$ tries got blue balls each time) then clearly this probability will be
$${p^k(1-p)^{n-k}}$$
where ${p}$ is the proportion of red balls to total number of balls (or if you like, the probability of getting a red ball in a single try).
But this isn't the only way of getting $k$ red balls. For example, I could have done
$${p^{k-1}p^{n-k}p}$$
That is, get ${k-1}$ red balls, then ${n-k}$ blue balls, then another red ball to finish. This is what the factor of
$${{n\choose k}}$$
keeps track of for you. It's a factor that tells you how many ways this could have occurred.
Hence
$${P\left(\text{getting exactly $k$ red balls}\right)={n\choose k}p^k(1-p)^{n-k}}$$
Now, we can simply plug this into our above sum. And we get
$${P\left(\text{getting at least $k$ red balls after $n$ tries}\right)=\sum_{i=k}^{n}P\left(\text{probability of getting exactly $i$ red balls}\right)}$$
$${=\sum_{i=k}^{n}{n\choose i}p^{i}(1-p)^{n-i}}$$
And you are done.

