Probability Ques 66
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
Show Answer
Answer:
Correct Answer: 66.(a,c)
Solution:
Formula:
Set theoretical notation of probability and some important results:
- 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.
BETA
