Let $B_0$ be the event that the batch has $0$ defectives, $B_1$ be the event the batch has $1$ defective, and $B_2$ be the event the batch has $2$ defectives.
Let $D_0$ be the event that neither selected component is defective. Problem (a) asks us to find the conditional probabilities $\Pr(B_0|D_0)$, $\Pr(B_1|D_0)$, and $\Pr(B_2|D_0)$. Now the hardest part, identifying precisely what we are after, has been done!
For the calculation, we use the general conditional probability formula
$$\Pr(X|Y)\Pr(Y)=\Pr(X\cap Y).$$
Put $X=B_0$ and $Y=D_0$. We need $\Pr(D_0)$ and $\Pr(B_0\cap D_0)$.
The event $D_0$ can happen in three different ways: (i) Our batch of $10$ is perfect, and we get no defectives in our sample of two; (ii) Our batch of $10$ has $1$ defective, but our sample of two misses them; (iii) Our batch has $2$ defective, but our sample misses them. If it helps, draw a tree that shows the three different paths through which we can end up with no defectives.
For (i), the probability is $(0.5)(1)$. For (ii), the probability that our batch has $1$ defective is $0.35$. Given that it has $1$ defective, the probability that our sample misses it is $\binom{9}{2}/\binom{10}{2}$, which is $8/10$. So the probability of (ii) is $(0.35)(8/10)$. For (iii), the probability our batch has $2$ defective is $0.15$. Given that it has $2$ defective, the probability that our sample misses both is $\binom{8}{2}/\binom{10}{2}$, which is $56/90$. So the probability of (iii) is $(0.15)(56/90)$. We have therefore found that
$$\Pr(D_0)=(0.5)(1)+(0.35)(8/10)+(0.15)(56/90).$$
The probability $\Pr(B_0\cap D_0)$ has been calculated during our calculation of $\Pr(D_0)$. It is $(0.5)(1)$. We conclude that
$$\Pr(B_0|D_0)=\frac{(0.5)(1)}{(0.5)(1)+(0.35)(8/10)+(0.15)(56/90)}.$$
The rest of the calculations for $D_0$ are easy, we have all the information needed. We get
$$\Pr(B_1|D_0)=\frac{(0.35)(8/10)}{(0.5)(1)+(0.35)(8/10)+(0.15)(56/90)}$$
and
$$\Pr(B_2|D_0)=\frac{(0.15)(56/90)}{(0.5)(1)+(0.35)(8/10)+(0.15)(56/90)}.$$
Alternately, we could have used Bayes' Formula directly . I wanted to do it in the above more basic way so that the logic would be clear.
Now unfortunately we have to deal with (b) and (c). But (c) is trivial! For (b), the calculation is as above, a little simpler, because if our sample of two has a defective, it cannot come from a perfect batch.
So in your comment you are saying that the lot is $100$, the number of defectives is $5$, and you seek $P(X\leq1)$
As the numbers keep on changing, the exact formula will be
$P(X\leq 1) = \dfrac{\binom 5 0\binom{95}{10}+\binom51\binom{95}9}{\binom{100}{10}} = \dfrac{43877}{47530}$
This is called a hypergeometric distribution.
Best Answer
The only way to get a binomial distribution is to assume that the sampling is done with replacement. If the sampling is without replacement, the distribution of the number of defectives is hypergeometric.
For your second question, if you pick $5$ items and number them, the probability the first is defective is $\frac{10}{25}$. The probability the second is defective is also $\frac{10}{25}$. And so on. Many people find this unintuitive at first.
Of course the conditional probability the second is defective, given the first is, is not $\frac{10}{25}$. When we do sampling without replacement, we lose independence.