Question

If $P\left(B|A\right)<P\left(B\right)$ the relationship between $P\left(A|B\right)$ and $P\left(A\right)$ is

• A. $P\left(A|B\right)<\frac{1}{2}P\left(A\right)$
• B. $P\left(A|B\right)>P\left(A\right)$
• C. $P\left(A|B\right)<P\left(A\right)$
• D. $P\left(A|B\right)<2P\left(A\right)$

Answer 1

The correct option is C $P\left(A|B\right)<P\left(A\right)$

First we note that
$P\left(\frac{B}{A}\right)=\frac{P(A\cap B)}{P\left(A\right)}<P\left(B\right)$ (given)
This means that
$\frac{P(A\cap B)}{P\left(A\right)}<P\left(B\right)$
So, $P(A\cap B)<P\left(A\right).P\left(B\right)$
divide throughout by P(B), we get
$\frac{P(A\cap B)}{P\left(B\right)}<P\left(A\right)$
Now ,we know that
$P\left(\frac{A}{B}\right)=\frac{P(A\cap B)}{P\left(B\right)}$
So, we conclude that
$P\left(\frac{A}{B}\right)<P\left(A\right)$
So, C is the correct answer.