Let $A\subset B$. Express the following probabilities as simply as possible:

$P(A\mid B),P\left(A\mid B^c\right),P(B\mid A),P\left(B\mid A^c\right)$

Assumption,

$A\subseteq B$

Note that the conditional probability of $P(X\mid Y)$is the proportion of $Y$that $X$makes.

This is the Venn diagram of the situation:

$P(A\mid B)$

$P(A\mid B)$should equal to the percentage of $B$that is in $A$. This is arbitrary defined by the size of $A$and $B$.

$A\subseteq B\Rightarrow AB=A$

The forms of $P(A\mid B)$

$P(A\mid B)=\frac{P\left(AB\right)}{P\left(B\right)}=\frac{P\left(A\right)}{P\left(B\right)}$

$P\left(A\mid B^c\right)$

Given that $B^c$happened, that is $B$did not happen, $A$also did not happen, therefore $P\left(A\mid B^c\right)=0$

$A\subseteq B\Rightarrow A{B}^c=\varnothing$

The forms of$P\left(A\mid B^c\right)$

$P\left(A\mid B^c\right)=\frac{P\left(A{B}^c\right)}{P\left(B^c\right)}=\frac{P(\varnothing )}{P\left(B^c\right)}=\frac{0}{P\left(B^c\right)}=0$

$P(B\mid A)$

Given that $A$happened, $B$certainly happened, therefore$P(B\mid A)=1$

$A\subseteq B\Rightarrow AB=A$

The forms of$P(B\mid A)$

$P(B\mid A)=\frac{P\left(AB\right)}{P\left(A\right)}=\frac{P\left(A\right)}{P\left(A\right)}=1$

$P\left(B\mid A^c\right)$

This is arbitrary defined by the size of $A$and $B$.

The forms of$P\left(B\mid A^c\right)$

$P\left(B\mid A^c\right)=\frac{P\left(A^{c}B\right)}{P\left(A^c\right)}=\frac{P\left(B\right)-P\left(A\right)}{1-P\left(A\right)}$

$P(A\mid B)=\frac{P\left(AB\right)}{P\left(B\right)}=\frac{P\left(A\right)}{P\left(B\right)}$

$P\left(A\mid B^c\right)=\frac{P\left(A{B}^c\right)}{P\left(B^c\right)}=\frac{P(\varnothing )}{P\left(B^c\right)}=\frac{0}{P\left(B^c\right)}=0$

$P(B\mid A)=\frac{P\left(AB\right)}{P\left(A\right)}=\frac{P\left(A\right)}{P\left(A\right)}=1$

$P\left(B\mid A^c\right)=\frac{P\left(A^{c}B\right)}{P\left(A^c\right)}=\frac{P\left(B\right)-P\left(A\right)}{1-P\left(A\right)}$