For two given events A and B, P,(A cap B) = | vedclass

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Question

Correct options are $A), B)$ and $C)$

$\because P(A \cup B)=P(A)+P(B)-P(A \cap B)$

on rearranging we get

$P(A \cap B)=P(A)+P(B)-P(A \cup B) \

Vedclass · Accepted Answer

All of the above

Vedclass · Answer

Correct options are $A), B)$ and $C)$

$\because P(A \cup B)=P(A)+P(B)-P(A \cap B)$

on rearranging we get

$P(A \cap B)=P(A)+P(B)-P(A \cup B) \quad$ which is option $C$

Also $\because \quad 0 \leq P ( A \cup B ) \leq 1$

$\Rightarrow P(A)+P(B)-1 \leq P(A \cap B) \leq P(A)+P(B)$

Hence $B$ and $C$ are also correct

Now if $P(A \cap B)=P(A)+P(B)+P(A \cap B)$

$\Rightarrow P ( A )+ P ( B )=0$

which is not necessary for the events $A$ and $B$

Hence the correct options are $A, B$ and $C$

For two given events $A$ and $B$, $P\,(A \cap B) = $

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