The equality of two conditional probability Suppose that a company has had $70\%$ female applicants and $30\%$ male applicants since it was founded. We also assume that $70\%$ of all staff are female. 
Let $F$ denote female and $M$ denote male and $S$ denote the event that an applicant is successful.
The question is: Is $P(S|F)=P(S|M)$?
Here is my working but perhaps it is not correct.
By the conditional probability we have
$$
\begin{align*}
P(S|F)&=\frac{P(S\cap F)}{P(F)}=\frac{0.7}{0.7}=1\\
P(S|M)&=\frac{P(S\cap M)}{P(M)}=\frac{0.3}{0.3}=1
\end{align*}
$$
So $P(S|F)=P(S|M)$.
 A: You're saying that given an applicant is female, they are guaranteed to get the job, and similarly for males. That is probably not true. $P(S\cap F)$ does not necessarily equal $0.7$. If there were $70$ female applicants of $100$ total applicants and $7$ females got the job then $P(S\cap F)=0.7\cdot0.1=0.07.$
We have
$$
\begin{align*}
P(S\mid F)&=\frac{P(F\mid S)\cdot P(S)}{P(F)}=\frac{0.7\cdot P(S)}{0.7}\\
P(S\mid M)&=\frac{P(M\mid S)\cdot P(S)}{P(M)}=\frac{0.3\cdot P(S)}{0.3}
\end{align*}
$$
From here the $P(S)$'s cancel and you can deduce the desired result.
A: Your mistake is claiming $\mathsf P(S\cap F)=70\%$.  We are not told that.


*

*We have been given that the proportion of applicants who are female is $70\%$, so we can estimate $\mathsf P(F)=70\%$.  

*We have also been given that the proportion of successful applicants who are female is $70\%$. This estimates $\mathsf P(F\mid S)=70\%$

*We do not know what $\mathsf P(S)$, the overall applicant success rate, may be, but $\mathsf P(F\mid S)=\mathsf P(F)$ immediately tells us something about the events $F$ and $S$. 


*

*What is this thing ?


*This in turn tells us something about $\mathsf P(S\mid F)$ and $\mathsf P(S\mid M)$, and their relation to $\mathsf P(S)$, whatever it is, and each other.   Indeed it tells us what we want: $\mathsf P(S\mid F)=\mathsf P(S\mid M)$.

A: Let $A$ and $C$ be the total numbers of applicants and current staff (so successful applicants), respectively. Then:
$$A=\underbrace{0.7A}_{F}+\underbrace{0.3A}_{M}; \ \ C=\underbrace{0.7C}_{F}+\underbrace{0.3C}_{M}.$$
Hence:
$$P(S|F)=\frac{P(F\cap S)}{P(F)}=\frac{0.7C}{0.7A}=\frac{C}{A}; \\ 
P(S|M)=\frac{P(M\cap S)}{P(M)}=\frac{0.3C}{0.3A}=\frac{C}{A}.$$
