$P(E|F) + P(E|F^{c})=1$ does it make sense? $P(E|F) + P(E|F^{c})=1$ does it make sense for any $E$ and $F$?
 A: No. Just take $E$ and $F$ (and $F^c$) independent and you'll get $2P[E]$.
On the other hand $P[E\mid F]+P[E^c\mid F]=1$ 
A: In general, no. $P(E|F) + P(E^C|F)=1$ makes sense, because:
$$P(E|F)+P(E^C|F)=\frac{P(E\cap F)}{P(F)}+\frac{P(E^C\cap F)}{P(F)}=\frac{P(F)}{P(F)}=1.$$
Example 1 (con): $100$ students took an exam. $80\%$ students had studied for the exam and $90\%$ of them passed it. $20\%$ students had not studied and $25\%$ of them passed it. 
Let $E=\{$passing exam$\}$, $F=\{$study for exam$\}$. Then:
$$P(E|F)=0.9; P(E|F^C)=0.25 \Rightarrow P(E|F)+P(E|F^C)=1.15>1;\\
=========================================\\
P(E|F)=0.9; P(E^C|F)=0.1 \Rightarrow P(E|F)+P(E^C|F)=1.$$
Example 2 (pro): $100$ students took an exam. $80\%$ students had studied for the exam and $90\%$ of them passed it. $20\%$ students had not studied and $10\%$ of them passed it. 
Let $E=\{$passing exam$\}$, $F=\{$study for exam$\}$. Then:
$$P(E|F)=0.9; P(E|F^C)=0.1 \Rightarrow P(E|F)+P(E|F^C)=1;\\
=========================================\\
P(E|F)=0.9; P(E^C|F)=0.1 \Rightarrow P(E|F)+P(E^C|F)=1.$$
