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14 May 2008 10:36:19 IST

Subject: Re:p n c

allamraju (3422)

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Olaaa!! Perrrfect answer.

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Every element x of the given set has four chances,They are

1)x $\epsilon$ P,x $\epsilon$ Q 2)x $\epsilon$ P,x doesnot belong to Q

3)x does not belong to P,x $\epsilon$ Q 4)x does not belong to both P,Q.

For P $\wedge$ Q to be $\phi$ ,the conditions 2,3,4 are favourable while 1 is not bcoz if x belongs to both sets,Their intersection can't be $\phi$ .So,the required probability is (3/4)ⁿ

MAKING A MISTAKE IS HUMAN BUT REPEATING IT IS IDIOTIC.

this reply: 2 points (with 0 Olaaa!! Perrrfect answer.

Olaaa!! Perrrfect answer.

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