Simple probability question, with faulty screws. I have translated the problem as follows:

A factory produces screws, the probability of them being faulty is 0.01 independently. The factory makes a box with 10 screws and recalls the boxes containing 2 or more faulty screws. What is the percentage of boxes that the factory has to recall?

 A: My solution is:
The percentage of faulty boxes is equal to the probability of faulty boxes.
In order to make easy my calculation, I calculate the probability of the boxes which have no faulty screws or have at least one faulty screw. Then I can find the probability I am searching for as follows:
$$S_{FaultyBox} = 1 - S_{NonFaultyBox}$$
To find the Non-Faulty Box I first find the probability that no box has faulty screws and then I sum to it the probability that it has only one faulty screw as follows:
$$S_{NoFaultyScrew} = (1-0.01)^{10}=0.904382$$
(Because they are independent of each other)
$$S_{OneFaultyScrew}= \binom{10}{1}0.01^{1}(1-0.01)^{9}=0.091351$$
(Using Bernoulli trials).
$$S_{NonfaultyBox} = S_{NofaultyScrew} + S_{OneFaultyScrew} = 0.995733$$
Finally, I find the probability that I was searching for:
$$S_{FaultyBox} = 1 - 0.995733 = 0.004267 \Rightarrow 0.4267\%$$
A: No faulty: $\big(\frac{9.99}{10}\big)^{10} = 0.9900\ldots$
$1$ faulty: $\big(\frac{9.9}{10}\big)^9\times \frac{0.01}{10}=0.0010\ldots$
So the probability is $\approx 0.9$%.
