$A= \{\text{an aircraft is present}\}$
$B= \{\text{the radar generates an alarm}\}$
$B^c=\{\text{the radar does not generate an alarm}\}$
$P(\text{present, no detection})=P(A)P(B^c|A)=0.05 \times 0.01=0.0005$
$$\text{Using Bayes' rule:}$$
$$P(\text{present, no detection})=P(A)P(B^c|A)=P(A)\frac{P(B^c)P(A|B^c)}{P(A)}=P(B^c)P(A|B^c)= (0.05 \times 0.01+ 0.95 \times 0.90)(0.01 + 0.90)≈0.78$$
Clearly, that's not right. So what am I doing wrong?
Data:
To be found are:
Calculations:
$P(A^{\complement}\cap B)=P(B\mid A^{\complement})P(A^{\complement})=P(B\mid A^{\complement})(1-P(A))=0.10\times(1-0.05)$
$P(A\cap B^{\complement})=P(B^{\complement}\mid A)P(A)=(1-P(B\mid A))P(A)=(1-0.99)\times0.05$
The last sequence of equalities in your question cannot be recognized as an application of Bayes rule. It is not more than just stating that: $$P(A)P(B^{\complement}\mid A)=P(A\cap B^{\complement})=P(B^{\complement})P(A\mid B^{\complement})\tag1$$
We can also find: $$P(B^{\complement})=1-P(B)=1-[P(B\mid A)P(A)+P(B\mid A^{\complement})P(A^{\complement})]=$$$$1-[P(B\mid A)P(A)+P(B\mid A^{\complement})(1-P(A)]=$$$$1-[0.99\times0.05+0.10\times(1-0.05)]$$
And combining this with $(1)$ we can find $P(A\mid B^{\complement})=\frac{P(A\cap B^{\complement})}{P(B^{\complement})}$.
But actually there is no need for finding $P(B^{\complement})$ and $P(A\mid B^{\complement})$
It is not clear to me how you found your expressions for $P(B^{\complement})$ and $P(A\mid B^{\complement})$
edit:
Your calculation of $P(B^{\complement})$ is clear now and correct. Both answers match.
The following is correct: $$\begin{align}P(B^c)&=0.05 \times 0.01+ 0.95 \times 0.90=0.8555.\\ (\color{red}{P(B^c)}&\color{red}{=P(A)\cdot P(B^c|A)+P(A^c)\cdot P(B^c|A^c)})\end{align}$$ The following is not correct: $$P(A|B^c)=0.01 + 0.90.\\ \left(\color{red}{P(A|B^c)=\frac{P(A\cap B^c)}{P(B^c)}=\frac{P(A)\cdot P(B^c|A)}{P(A)\cdot P(B^c|A)+P(A^c)\cdot P(B^c|A^c)}=\frac{0.05\cdot 0.01}{0.8555}\approx0.00058}\right).$$ which is actually circular.
Finally, Eureka!
I thought that $P(A|B^c)=0.05 \times 0.01$ because it's the only case that matters. Like, I want to know what's the probability of presence of an aircraft and absence of the alarm at the same time; and I think, here it is: $P=0.05$ that there's an aircraft times $P=0.01$ that there's no alarm given that an aircraft is present — done, case closed, that's the exact combination I need. But this combination relates to the knowledge of the presence, which I don't have. So I wrongly assume — as I emphasized with bold above — that I should only care about the presence and don't reckon with the other effects(which in this problem is $P(B^c|A^c)$). And the theory distinctly says $$P(A|B)=\frac{P(A \cap B)}{P(B)}$$
With that said, it should be like this: $$P(\text{present, no detection})=P(B^c)P(A|B^c)=[P(B^c|A)+P(B^c|A^c)]\times \frac{P(B^c|A)}{P(B^c|A)+P(B^c|A^c)}=P(B^c|A)=0.05 \times 0.01$$
Imagine repeating this scenario 10000 times. "We assume an aircraft is present with probability 0.05". So in 10000*0.05= 500 cases an aircraft is present. In the other 9500 cases it is not. "If an aircraft is present in an area a radar detects it 99% of the time". So of the 500 cases where an aircraft is present, radar detects it 500*0.99= 495 times but fails to detect it 5 times. "If an aircraft is not present, the radar generates a (false) alarm with probability 0.10". Of the 9500 cases where an airplane is not present, there is a false alarm 9500*0.10= 950 times.
"What is the probability of no aircraft present and a false alarm?" There were 950 occurrences of "no airplane present and a false alarm" in the 10000 scenarios so the probability of "no airplane present and a false alarm is 950/10000= 0.095.
"What is the probability of airplane presence and no alarm?" There were 5 times when an airplane was present and there was no alarm in the 10000 scenarios so the probability of "an airplane present and no alarm" is 5/10000= 0.0005.
