How do you calculate $P(X = nG), P(X = nG^C)$ by the Law of Total Probability, with extra conditioning?

Please see $P(UG)$ and $P(UG^C)$ below, beside my red line. Can you please expound these calculations?

How was $s$ computed in both equations?
Example 2.4.5 (Unanimous agreement).
The article "Why too much evidence can be a bad thing" by Lisa Zyga [30] says:
Under ancient Jewish law, if a suspect on trial was unanimously found guilty by all judges, then the suspect was acquitted. This reasoning sounds counterintu itive, but the legislators of the time had noticed that unanimous agreement often indicates the presence of systemic error in the judicial process.
There are n judges deciding a case. The suspect has prior probability p of being guilty. Each judge votes whether to convict or acquit the suspect. With probability s, a systemic error occurs (e.g., the defense is incompetent). If a systemic error occurs, then the judges unanimously vote to convict (i.e., all n judges vote to convict).
Whether a systemic error occurs is independent of whether the suspect is guilty. Given that a systemic error doesn't occur and that the suspect is guilty, each judge has probability c of voting to convict, independently. Given that a systemic error doesn't occur and that the suspect is not guilty, each judge has probability w of voting to convict, independently. Suppose that
$0 < p < 1; 0 < s < 1;$ and $0 < w < \frac12 < c < 1$.
(a) For this part only, suppose that exactly k out of n judges vote to convict, where k < n. Given this information, find the probability that the suspect is guilty.
(b) Now suppose that all n judges vote to convict. Given this information, find the probability that the suspect is guilty.
(c) Is the answer to (b), viewed as a function of n, an increasing function? Give a short, intuitive explanation in words.
Solution: (a) Since $k<n$, a systemic error didn't occur. We will implicitly condition on this in this part. Let $G$ be the event that the suspect is guilty and $X$ be the number of judges who vote to convict. Using Bayes' rule, LOTP, and the Binomial PMF, $$ P(G \mid X=k)=\frac{P(X=k \mid G) P(G)}{P(X=k)}=\frac{p c^{k}(1c)^{nk}}{p c^{k}(1c)^{nk}+(1p) w^{k}(1w)^{nk}} $$ (b) Let $U$ be the event $X=n$ and $B$ be the event that a systemic error occurs. Then $$ P(G \mid U)=\frac{P(U \mid G) P(G)}{P(U)}=\frac{p P(U \mid G)}{p P(U \mid G)+(1p) P\left(U \mid G^{c}\right)} $$ By LOTP with extra conditioning,
$$ \color{red}{\begin{aligned} P(U \mid G) &=P(U \mid G, B) P(B \mid G)+P\left(U \mid G, B^{c}\right) P\left(B^{c} \mid G\right)=s+(1s) c^{n} \\ P\left(U \mid G^{c}\right) &=P\left(U \mid G^{c}, B\right) P\left(B \mid G^{c}\right)+P\left(U \mid G^{c}, B^{c}\right) P\left(B^{c} \mid G^{c}\right)=s+(1s) w^{n} \end{aligned}} $$
Thus, $$ P(G \mid U)=\frac{p\left(s+(1s) c^{n}\right)}{p\left(s+(1s) c^{n}\right)+(1p)\left(s+(1s) w^{n}\right)} $$ (c) No, since a large value of $n$ yields a high chance of systemic error, and if a systemic error occurs then the judges' votes are uninformative about whether the suspect is guilty. The answer to (b) reverts to the prior probability $p$ as $n \rightarrow \infty$
Blitzstein, Introduction to Probability (2019 2 ed), Example 2.4.5, p 62.
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