Probability 2 Question 5
5. If $0<P(A)<1,0<P(B)<1$ and $P(A \cup B)=P(A)$ $+P(B)-P(A) P(B)$, then
$(1995,2 M)$
(a) $P(B / A)=P(B)-P(A)$
(b) $P\left(A^{\prime}-B^{\prime}\right)=P\left(A^{\prime}\right)-P\left(B^{\prime}\right)$
(c) $P(A \cup B)^{\prime}=P(A)^{\prime} P(B)$
(d) $P(A / B)=P(A)-P(B)$
Show Answer
Answer:
Correct Answer: 5. (a)
Solution:
- Since, $P(A \cap B)=P(A) \cdot P(B)$
It means $A$ and $B$ are independent events, so $A^{\prime}$ and $B^{\prime}$ are also independent.
$\therefore \quad P(A \cup B)^{\prime}=P\left(A^{\prime} \cap B^{\prime}\right)=P(A)^{\prime} \cdot P(B)^{\prime}$
Alternate Solution
$$ \begin{aligned} P(A \cup B)^{\prime} & =1-P(A \cup B)=1-{P(A)+P(B)-P(A) \cdot P(B)} \\ & ={1-P(A)}{1-P(B)}=P(A)^{\prime} P(B)^{\prime} \end{aligned} $$
