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If $P(A)=0.8, P(B)=0.9, P(A B)=p$, which one ofthe following is correct?
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Verified Answer
The correct answer is:
$\quad 0.7 \leq p \leq 0.8$
We know,
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow 0.8+0.9-p \leq 1$
$\Rightarrow \quad 1.7-p \leq 1$
$\Rightarrow \quad 0.7 \leq p$
Now, $\quad P(A) < P(B)$
$\therefore \quad P(A \cap B) \leq P(A)$
$\Rightarrow \quad p \leq 0.8$
Hence, $0.7 \leq p \leq 0.8$
$P(A \cup B)=P(A)+P(B)-P(A \cap B)$
$\Rightarrow 0.8+0.9-p \leq 1$
$\Rightarrow \quad 1.7-p \leq 1$
$\Rightarrow \quad 0.7 \leq p$
Now, $\quad P(A) < P(B)$
$\therefore \quad P(A \cap B) \leq P(A)$
$\Rightarrow \quad p \leq 0.8$
Hence, $0.7 \leq p \leq 0.8$
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