Q. 3.2 - A First Course in Probability

Question

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

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

Step-by-Step Solution

Answer

$P(A\mid B), 0, 1, P\left(B\mid A^c\right)$

$A\subset B$ means that if $A$ occurs, $B$ will too, that is that there is no $A$ in $B^c$.

Step 1: Given Information

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:

Step 2: Explanation

$P(A\mid B)$

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

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

Step 3: Final Answer

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

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

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

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