Probability of drawing a total of at least x amount of black balls from different boxes?

Question

Suppose we have two boxes containing a 100 balls. Box A has 45 black, 55 white balls and Box B has 30 black and 70 white balls. There is no trick with the boxes and the drawing process is completely random. Every time a ball is picked, it will be dropped back into the box. A person will pick a ball from Box A 23 times and 18 times from Box B. What is the probability that he picks a black ball at least 30 times?

The answer I came up with if the question asked the probability of drawing a black ball exactly 30 times is this (it may very well be wrong):

$\sum_{n=12}^{23}\binom{23}{n}(0.45)^n(0.55)^{23-n} \binom{18}{30-n}(0.3)^{30-n}(0.7)^{n-12}$

Now, I'm sure there is a better way of doing this for every number greater than 30 until I reach 41 and then add all of those probabilities. I'm drawing a blank. Any help will be appreciated.

Answer 1

The answer I came up with if the question asked the probability of drawing a black ball exactly 30 times is this (it may very well be wrong):

$\sum_{n=12}^{23}\binom{23}{n}(0.45)^n(0.55)^{30-n} \binom{18}{30-n}(0.3)^{30-n}(0.7)^{n-12}$

It is wrong, but very close (possibly a typo).

$$\sum_{n=12}^{23}\binom{23}{n}(0.45)^n(0.55)^{\mathbf{23}-n} \binom{18}{30-n}(0.3)^{30-n}(0.7)^{n-12}$$

Now, I'm sure there is a better way of doing this for every number greater than 30 until I reach 41 and then add all of those probabilities. I'm drawing a blank. Any help will be appreciated.

Sadly, nope. $\ddot\frown$

$$\sum_{m=30}^{41}\sum_{n=m-18}^{23}\binom{23}{n}(0.45)^n(0.55)^{\mathbf{23}-n} \binom{18}{m-n}(0.3)^{m-n}(0.7)^{18-m+n}$$