Probability 3 Question 27
27. For any two events $A$ and $B$ in a sample space
(1991, 2M)
(a) $P \frac{A}{B} \geq \frac{P(A)+P(B)-1}{P(B)}, P(B) \neq 0$ is always true
(b) $P(A \cap \bar{B})=P(A)-P(A \cap B)$ does not hold
(c) $P(A \cup B)=1-P(\bar{A}) P(\bar{B})$, if $A$ and $B$ are independent
(d) $P(A \cup B)=1-P(\bar{A}) P(\bar{B})$, if $A$ and $B$ are disjoint
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Solution:
- We know that,
$$ P \frac{A}{B}=\frac{P(A \cap B)}{P(B)}=\frac{P(A)+P(B)-P(A \cup B)}{P(B)} $$
$$ \begin{aligned} & \text { Since, } \quad P(A \cup B)<1 \\ & \Rightarrow \quad-P(A \cup B)>-1 \\ & \Rightarrow \quad P(A)+P(B)-P(A \cup B)>P(A)+P(B)-1 \\ & \Rightarrow \quad \frac{P(A)+P(B)-P(A \cup B)}{P(B)}>\frac{P(A)+P(B)-1}{P(B)} \\ & \Rightarrow \quad P \frac{A}{B}>\frac{P(A)+P(B)-1}{P(B)} \end{aligned} $$
Hence, option (a) is correct.
The choice (b) holds only for disjoint i.e. $P(A \cap B)=0$
Finally, $P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$$ =P(A)+P(B)-P(A) \cdot P(B), $$
if $A, B$ are independent
$$ =1-{1-P(A)}{1-P(B)}=1-P(\bar{A}) \cdot P(\bar{B}) $$
Hence, option (c) is correct, but option (d) is not correct.
