The question asks you to find the probability of getting $2$ defective in the sample of $25$ if the true proportion of output that is defective is $1$%, that is, $0.01$. So the correct answer is
$$\binom{25}{2}(0.01)^2(0.99)^{23}.$$
This seems to be roughly $0.0238$.
Part ii) has to do with hypothesis testing. Roughly speaking, we have the Null Hypothesis that indeed the supplier's output is only $1$% defective. We have performed an experiment. Even if the supplier's output is only $1$% defective, by bad luck there could be $2$ defectives in the batch of $25$ that was tested.
But this would not happen very often. If the Null Hypothesis holds, then only about $2.38$% of batches of $25$ would have exactly $2$ defectives. So on the "$1$%" hypothesis, something rather unusual has happened. Not impossible, but unusual.
In hypothesis testing, it is common to set a significance level, moderately often $5$%, or $1$%. If in testing something happens which (under the Null Hypothesis) would happen with probability less than $0.05$, then at significance level $0.05$, one rejects the Null Hypothesis. So at significance level $0.05$, the experimental result described is enough to reject the Null Hypothesis.
By the way, a real statistician would calculate the probability of getting $\ge 2$ defective under the Null Hypothesis. This is about $0.02576$. Still well under $0.05$. I am somewhat troubled that the problem setter focused on exactly $2$, this is the wrong approach to hypothesis testing.
Further comments: To me, using level of significance $0.05$ seems unsuitable, I would want stronger evidence to reject the Null Hypothesis. After all, if you have a long-established relationship with a supplier, it seems foolish to dump it on weakish evidence. Also, it is hard to select $25$ items really randomly. Even if overall output is only $1$% defective, there can be non-random day to day variation.
More sensible is to consider the test results as a warning that something might be wrong. So I would at least recommend that more testing be done, since the Null Hypothesis certainly did not get a clean bill of health.
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.
Best Answer
(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).