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.
Your description of $A$ and $B$ isn’t quite right: you want $A$ to be the event that Component $1$ is selected (which you don’t actually need), and $B$ to be the event that Component $2$ is selected.
Your calculation of $P(F)$ is fine. $P(B\cap F)$ is the probability that you picked Component $2$ and Component $2$ is faulty. The events $B$ and $F$ are independent, so this $P(B)\cdot P(F)$. We’re told that the component was picked ‘at random’, which is a sloppy way of saying that each component had probability $\frac12$ of being picked. Thus, $P(B)\cdot P(F)=\frac12\cdot\frac1{10}=\frac1{20}$. Thus, your computation used the correct value of $P(B\cap F)$ (and came to the correct final result)but contrary to what you then said, this is not $P(B)$: $P(B)$ is $\frac12$.
You could also use Bayes’ theorem for this problem:
$$P(B\mid F)=\frac{P(F\mid B)\cdot P(B)}{P(F)}\;.$$
We already know that $P(F)=\frac3{20}$ and $P(B)=\frac12$, and $P(F\mid B)$ is simply the probability that Component $2$ is faulty given that it was picked. Whether or not Component $2$ is faulty does not depend on whether it was picked, so this is simply $\frac1{10}$.
Best Answer
You're almost there! You've made a small mistake though, which even @antkam forgot to mention. With your method, using the complement won't yield the right answer, since, if we let X be the number of components which failed QA tests;
So by that logic, the probability you're looking for is $P(x\leq1)=P(x=0)+P(x=1)$, and NOT $1-P(x=1)$.
Merry Christmas! :)