Conditional Probability with two subsets

Question

Question:

A man plans to ship six boxes. Two of the boxes are insured, while the other four aren't. Each package that is shipped has a 10% chance of being damaged.

What is the probability that:

1. neither of the insured packages were damaged, given that exactly 3 packages were damaged?

   2. exactly three packages were damaged, given that neither of the two insured packages were damaged?

I know this is a Bayesian question, where the two insured packages would be $(.9)^2 = (.82)$.

For Question 1:

$\large \frac{{4 \choose 3}(.1)^3(.9)^3}{{6 \choose 3}(.1)^3(.9)^3}$

Just want to make sure I'm doing this correctly.

Answer 1

2) Let $ A = \{ \text{exactly 3 were damaged}\}$ and $B = \{\text{neither of the insured were damaged}\}$ \begin{align} &P(A|B) = \frac{P(A\cap B)}{P(B)} = \frac{\binom{4}{3}0.1^30.9^3}{0.9^2} = \binom{4}{3}0.1^30.9 \end{align}