MCQ
If $\text{P}\text{(A}\cup\text{B)} = \text{P}\text{(A}\cap\text{B)}$ for any two events A and B, the
- A$\text{P(A)}=\text{P(B)}$
- B$\text{P(A)}>\text{P(B)}$
- C$\text{P(A)}<\text{P(B)}$
- DNone of these.
If $\text{P}\text{(A}\cup\text{B)} = \text{P}\text{(A}\cap\text{B)}$ for any two events A and B, the
We know that,
$\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{(B})-\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A)}+\text{P}\text{(B})-\text{P}(\text{A}\cap\text{B})$ $\big[\text{P}(\text{A}\cup\text{B})=\text{P}(\text{A}\cup\text{B})\big]$
$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})=0\ ...(1)$
But,
$\text{P(A)}-\text{P}(\text{A }\cap\text{ B})\geq0$
$\text{P(B)}-\text{P}(\text{A }\cap\text{ B})\geq0$
$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})\geq0\ ...(2)$
From (1) and (2), we have,
$\Rightarrow\text{P}(\text{A)}-\text{P}(\text{A}\cup\text{B})+\text{P}(\text{B)}-\text{P}(\text{A}\cap\text{B})=0$
$\Rightarrow\text{P}(\text{A)}=\text{P}(\text{A}\cap\text{B})\text{ and }\text{P}(\text{B)}=\text{P}(\text{A}\cap\text{B})$
$\Rightarrow\text{P(A)}=\text{P(B)}$
Hence, the correct answer is option (a).
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