Help with a probability homework problem I'm trying to solve the problem in the image but having trouble formulating the problem mathematically:

There are $20$ identical laptops on a trolley, out of which $12$ have
a hard disk with capacity $160$ GB and $8$ with a capacity of $320$
GB. A teacher randomly takes two laptops from the trolley. A student
then takes a laptop from the trolley to complete a project. Given that
the student took a laptop with $160$ GB, find the probability that the
teacher took both laptops with $320$ GB.

so any help will be appreciated.
I've tried the following: let $A$ denote the event of choosing a 160GB laptop, and $B$ denote the event of choosing a 320GB laptop. Then $P(A)= 12/20, P(B)=8/20.$ Next, let $C$ denote the event of choosing 2 laptops, regardless of their memory (160GB or 320GB). Then $P(C)=2/20$ (right?) Then $B \cap C$ will denote the event of choosing two 320GB laptops (right?). But then how do I calculate $P(B \cap C)?$
Also, how do I calculate the probability of $D:=$  choosing two 160GB laptops and $P(A|D)$? I understand that at the end we'll need to calculate $P(D|A)$, and hence we'll need to use the formula $P(D|A)= \frac{P(D \cap A)}{P(A)}= 20/12 * P(D \cap A)=20/12 * P(A|D) P(D).$ Here's where I'm confused: how to I calculate $ P(A|D), P(D)?$
Let the variable $L_1$ be $0$ or $1$ depending on whether the first laptop taken is small or big.
Let the variable $L_2$ be $0$ or $1$ depending on whether the second laptop taken is small or big.
Let the variable $L_3$ be $0$ or $1$ depending on whether the third laptop taken small or big.
The question is to compute $P(L_1 = 1, \, L_2 = 1 \,|\, L_3 = 0)$.
$$P(L_1 = 1, \, L_2 = 1 \,|\, L_3 = 0) = P(L_1 = 1, \, L_2 = 1, \, L_3 = 0) / P(L_3 = 0) $$
You can probably figure out how to compute the numerator of the right hand side. To compute the denominator, note that the three laptops chosen are a uniform random trio. Therefore each laptop in the chosen trio has the same claim to be small or big as the other two in the trio. Therefore $P(L_3 = 0) = P(L_1 = 0) = 12/20$.
That the three laptops chosen are a random trio arises from a more general phenomenon. If you consider all size $N$ subsets of a set, and choose one of those subsets uniformly at random, it's the same as if you chose $N$ elements one a time uniformly. You can show this using induction.
I don't think the student picking a laptop with a 160 GB hard drive after the fact affects the probability of the teacher choosing two laptops with 320 GB hard drives. If the question was what is the probability of the teacher choosing two 320 GB laptops and then a student choosing a 160 GB, that would be a different story. But, let's focus on the question at hand.
Let A represent the event of choosing a laptop with a 320 GB hard drive on the first try and B represent the event of choosing a laptop with a 320 GB hard drive on the second try.
The probability of choosing 320 GB on the first try and 320 GB on the second try is $$P(A\cap B) = P(B|A)P(A).$$
The probability of A is calculated as follows:
$$P(A) = \frac{8}{20} = \frac{2}{5}.$$
Since there are 7 laptops with 320 GB hard drives remaining of 19 total laptops, the probability of B given A is calculated as follows:
$$P(B|A) = \frac{7}{19}$$.
Therefore, the probability of both A and B occurring is:
$$P(A\cap B) = P(B|A)P(A) = (\frac{7}{19})(\frac{2}{5}) = \frac{14}{95}.$$