Question

If $A$ and $B$ are two events such that $A\subset B$ and $P\left(B\right)\ne 0$, then which of the following is correct ?

• A. $P\left(A|B\right)=\frac{P\left(B\right)}{P\left(A\right)}$
• B. $P\left(A|B\right)<P\left(A\right)$
• C. $P\left(A|B\right)\ge P\left(A\right)$
• D. $Noneofthese$

Answer 1

Answer: (A), (C)

The correct options are
A $P\left(A|B\right)=\frac{P\left(B\right)}{P\left(A\right)}$
C $P\left(A|B\right)\ge P\left(A\right)$

If $A\subset B$, then $A\cap B=A$
$\Rightarrow P(A\cap B)=P\left(A\right)$
Also, $P\left(A\right)<P\left(B\right)$
Consider $P\left(A|B\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{P\left(A\right)}{P\left(B\right)}\ne \frac{P\left(B\right)}{P\left(A\right)}$ .....(1)
Consider $P\left(A|B\right)=\frac{P(A\cap B)}{P\left(B\right)}=\frac{P\left(A\right)}{P\left(B\right)}$ ....(2)
It is known that, $P\left(B\right)\le 1$
$\Rightarrow \frac{1}{P\left(B\right)}\ge 1$
$\Rightarrow \frac{P\left(A\right)}{P\left(B\right)}\ge P\left(A\right)$
From (2), we obtain
$\Rightarrow P\left(A|B\right)\ge P\left(A\right)$ ....(3)
$\therefore P\left(A|B\right)$ is not less than $P\left(A\right)$.
Thus, from (3), it can be concluded that the relation given in alternative C is correct.