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  4. If n(A) = 1000, n(B) = 500, n(A ∩ B) ge 1 and n(A ∪ B) = p, then

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Q. If $n(A) = 1000$, $n(B) = 500$, $n(A \cap B) \ge 1$ and $n(A \cup B) = p$, then

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Solution:

We know that, $n(A \cup B) = n(A) + n(B) - n(A \cap B)$
$\Rightarrow p = 1000 + 500 - n (A \cap B)$ $\Rightarrow 1 \le n(A \cap B) \le 500$
Hence, $p = 1000 + 500 - 1 = 1499$
and $p = 1000 + 500-500= 1000$
$\therefore 1000 \le p \le 1499$