Probability Ques 74
If $A$ and $B$ are two independent events, prove that $P(A \cup B) \cdot P\left(A^{\prime} \cap B^{\prime}\right) \leq P(C)$, where $C$ is an event defined that exactly one of $A$ and $B$ occurs. $\quad$
(2004, 2M)
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Solution:
Formula:
Set theoretical notation of probability and some important results:
- Here, $P(A \cup B) \cdot P\left(A^{\prime} \cap B^{\prime}\right)$
$ \Rightarrow \{P(A)+P(B)-P(A \cap B)\}\{P\left(A^{\prime}\right) \cdot P\left(B^{\prime}\right) \} $
[since $A, B$ are independent, so $A^{\prime}, B^{\prime}$ are independent]
$\therefore P(A \cup B) \cdot P\left(A^{\prime} \cap B^{\prime}\right) \leq\{P(A)+P(B)\} \cdot\{P\left(A^{\prime}\right) \cdot P(B) \}$
$ =P(A) \cdot P\left(A^{\prime}\right) \cdot P(B)+P(B) \cdot P\left(A^{\prime}\right) \cdot P(B) $
$ \leq P(A) \cdot P(B)+P(B) \cdot P\left(A^{\prime}\right) $
$ \left[\because P\left(A^{\prime}\right) \leq 1 \text { and } P\left(B^{\prime}\right) \leq 1\right] $
$\Rightarrow P(A \cup B) \cdot P\left(A^{\prime} \cap B^{\prime}\right) \leq P(A) \cdot P\left(B^{\prime}\right)+P(B) \cdot P\left(A^{\prime}\right)$
$\Rightarrow P(A \cup B) \cdot P\left(A^{\prime} \cap B^{\prime}\right) \leq P(C)$
$ \left[\because P(C)=P(A) \cdot P\left(B^{\prime}\right)+P(B) \cdot P\left(A^{\prime}\right)\right] $
BETA
