Question

Let $0<P\left(A\right)<1,0<P\left(B\right)<1$ and $P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)$, then

• A. $P(B\cap A^{\prime })=P\left(B\right)-P\left(A\right)$
• B. $P(A^{\prime }\cup B^{\prime })=P\left(A^{\prime }\right)+P\left(B^{\prime }\right)$
• C. $P{(A\cup B)}^{\prime }=P\left(A^{\prime }\right)P\left(B^{\prime }\right)$
• D. $P\left(\frac{A}{B}\right)=P\left(A\right)$

Answer 1

Answer: (C), (D)

$P{(A\cup B)}^{\prime }=P\left(A^{\prime }\right)P\left(B^{\prime }\right)$
$P\left(\frac{A}{B}\right)=P\left(A\right)$

$P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)\Rightarrow P\left(A\right)+P\left(B\right)-P(A\cap B)=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)\Rightarrow P(A\cap B)=P\left(A\right)P\left(B\right)$
$\therefore A$ and $B$ are independent events.
$P(B\cap A^{\prime })=P\left(B\right)P\left(A^{\prime }\right)\ne P\left(B\right)-P\left(A\right)$
and $P(A^{\prime }\cup B^{\prime })=P(A\cap B^{\prime })=1-P(A\cap B)=1-P\left(A\right)P\left(B\right)$
$\ne 1-P\left(A\right)+1-P\left(B\right)=P\left(A^{\prime }\right)+P\left(B^{\prime }\right)$
Also $P{(A\cup B)}^{\prime }=P(A^{\prime }\cap B^{\prime })=P\left(A^{\prime }\right)P\left(B^{\prime }\right)$
Since $A$ and $B$ are independent $P\left(\frac{A}{B}\right)=P\left(A\right)$