Question
If P(A) = 0.4, P(B) = 0.8, $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.6$. Find $\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)$ and $\text{P}(\text{A}\cap\text{B}).$

Answer

Given:
P(A) = 0.4, P(B) = 0.8, $\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=0.6$
We know that,
$\text{P}\Big(\frac{\text{B}}{\text{A}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(A)}}$
$0.6=\frac{\text{P}(\text{A}\cap\text{P})}{0.4}$
$\text{P}(\text{A}\cap\text{B})=0.6\times0.4$
$\text{P}(\text{A}\cap\text{B})=0.24$
Now,
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=\frac{\text{P}(\text{A}\cap\text{B})}{\text{P(B)}}$
$=\frac{0.24}{0.8}$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.3$
$\text{P}(\text{A}\cap\text{B})=\text{P(A)}+\text{P(B)}-\text{P}(\text{A}\cap\text{B})$
$=0.4+0.8-0.24$
$\text{P}(\text{A}\cap\text{B})=0.96$
$\text{P}\Big(\frac{\text{A}}{\text{B}}\Big)=0.3, \text{P}(\text{A}\cap\text{B})=0.96$

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