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

Answer 1

Answer: (C), (D)

The correct options are
C $P\left(\right(A\cup B{)}^C)=P(A^C\left)P\right(B^C)$
D $P\left(\frac{A}{B}\right)=P\left(A\right)$

Given,
$P(A\cup B)=P\left(A\right)+P\left(B\right)-P\left(A\right)P\left(B\right)$
1)$P\left(\frac{B}{A}\right)=P\left(\frac{P(B\cap A)}{A}\right)$$=P\left(\frac{P\left(B\right)\ast P\left(A\right)}{A}\right)=P\left(B\right)$
2)$P(A^c\cup B^c)=P\left(A^c\right)+P\left(B^c\right)-P(A^c\cap B^c)$
3)$P\left(\right(A\cup B{)}^c)=(1-P(A\cup B))=(1-P\left(A\right)-P\left(B\right)+P\left(A\right)P\left(B\right))$
$\Rightarrow P\left(\right(A\cup B{)}^c)=(1-P\left(A\right)\left)\right(1-P\left(B\right))=P(A^c\left)P\right(B^c).$
4)$P\left(\frac{A}{B}\right)=\left(\frac{P(A\cap B)}{P\left(B\right)}\right)$
$P\left(\frac{A}{B}\right)=\left(\frac{P\left(A\right)P\left(B\right)}{B}\right)=P\left(A\right)$
Therefore,options 3,4 are correct