Question

If $A$ and $B$ are two events such that $2P\left(A\right)=3P\left(B\right)$, where $0<P\left(A\right)<P\left(B\right)<1$, then which one of the following is correct?

• A. $P\left(A|B\right)<P\left(B|A\right)<P(A\cap B)$
• B. $P(A\cap B)<P\left(B|A\right)<P\left(A|B\right)$
• C. $P\left(B|A\right)<P\left(A|B\right)<P(A\cap B)$
• D. $P(A\cap B)<P\left(A|B\right)<P\left(B|A\right)$

Answer 1

Answer: (B)

$P(A\cap B)<P\left(B|A\right)<P\left(A|B\right)$

Given$:-$

$2P\left(A\right)=3P\left(B\right)$

$\Rightarrow$ $P\left(A\right)=\frac{3P\left(B\right)}{2}$ $\to ⟨1⟩$

$\Rightarrow \therefore P\left(A\right)>P\left(B\right)$

Now we know$:-$

$P\left(A|B\right)=\frac{P(A\cap B)}{P\left(B\right)}$

Similarly,

$P\left(B|A\right)=\frac{P(A\cap B)}{P\left(A\right)}$

$\Rightarrow P\left(B|A\right)=\frac{2P(A\cap B)}{3P\left(B\right)}$ $\left[Using⟨1⟩\right]$

$\therefore$ From above we conclude $P\left(A|B\right)>P\left(B|A\right)$

Now also we know, $0<P\left(A\right)<1$ and $0<P\left(B\right)<1$

$\therefore P\left(B|A\right)=\frac{P(A\cap B)}{P\left(A\right)}>P(A\cap B)$

Similarly,

$P\left(A|B\right)=\frac{P(A\cap B)}{P\left(B\right)}>P(A\cap B)$

$\therefore P\left(A|B\right)>P\left(B|A\right)>P(A\cap B)$