(a) Imagine that two components are selected, one at a time, from a group of $10$ that contains $2$ bad. The probability the first item selected is good is $\frac{8}{10}$. Given that the first item was good, the probability the second item is good is $\frac{7}{9}$. So our probability is $\frac{8}{10}\cdot\frac{7}{9}$.
Another way: There are $\binom{10}{2}$ equally likely ways to select $2$ items. There are $\binom{8}{2}$ ways to select $2$ good. For the probability, divide.
(b) The probability a batch with $2$ bad is selected is $0.2$. Given this has happened, the probability both items tested are good is the answer to (a)$. Multiply.
(c) This will require some calculation. We get that the tested items are both good in $3$ different ways: (i) The batch of $10$ chosen is all good or (ii) the batch chosen has $1$ bad, but it is not one of the tested items or (ii) the batch chosen has $2$ bad, neither of which is tested. We find the probabilities of (i), (ii), and (iii), and add up.
In (b), we found the probability of (iii).
If the items are all good, then for sure the $2$ tested items will be good. Thus the probability of (i) is $(0.5)(1)$.
To find the probability of (ii), use the method that we used to solve (a) and (b).
Let $A$ be the event the parts were selected from the first shipment, and let $G$ be the event they are both good.
We want the probability they are from the first shipment, given they are both good. So we want $\Pr(A|G)$.
By the definition of conditional probability, we have
$$\Pr(A|G)=\frac{\Pr(A\cap G)}{\Pr(G)}.\tag{1}$$
We find the two probabilities on the right-hand side of (1).
The event $G$ can happen in two disjoint ways: (i) we select from the first shipment, and both items are good or (ii) we select from the second shipment, and both items are good.
We find the probability of (i). The probability we choose from the first shipment is $\frac{1}{2}$. Given that we selected from the first shipment, the probability they were both good is $\frac{900}{1000}\cdot \frac{899}{999}$.
Thus the probability of (i) is $\dfrac{1}{2}\cdot\dfrac{900}{1000}\cdot\dfrac{899}{999}$.
Similarly, find the probability of (ii). Add to get $\Pr(G)$.
Note that the numerator in Formula (1) is just what we called the probability of (i).
Remark: We interpreted "$10\%$ bad" literally, as in exactly $10$ percent, that is, exactly $100$ bad items in the group of $1000$. However, another reasonable interpretation is that the first shipment comes from a supplier who has a $10\%$ bad rate. Then we would replace $\frac{900}{1000}\cdot \frac{899}{999}$ by $\left(\frac{90}{100}\right)^2$. Numerically, it makes no practical difference.
Best Answer
For $Y=2$: (both defectives are found in the first two inspections) $$P(Y=2)=\text{(prob. of first defective) } \cdot \text{(prob. of second defective) }=\frac{2}{4} \cdot \frac{1}{3}.$$ The reason is: the prob.e of finding the first defective is $2$ out of total of $4$ components. Now once that is found there is only one defective left, hence the prob. of finding the second one now is $1$ out of a total of $3$.
For $Y=3$. You need to break it into two cases: $GDD$ and $DGD$. Where $GDD$ means the first component tested was "G"ood but the second was "D"efective and the third was "D"efective as well. You can guess what $DGD$ means. Observe that there cannot be $DDG$ because then the test would have ended with two trials only.
So $$P(Y=3)=\underbrace{\binom{2}{1}}_{\text{choosing how to place $G$}}\quad \underbrace{\left(\frac{2}{4}\cdot\frac{2}{3}\right)}_{\text{prob of $GD$}} \quad \cdot \quad \underbrace{\frac{1}{2}}_{\text{prob. of having the last component $D$}}.$$
I hope this will help.
NOTE:
For $Y=4$: you need to have the last tested component be $D$ and the first three should have 2$G's$ and 1$D$.
choosing a place for the first $D$: $\binom{3}{1}$, now fill in the details..