A room has 4 doors (D₁, D₂, D₃, D₄). A person entering through a door can't leave by the same door and is equally likely to leave from the rest. A person P enters the room and leaves. Then which of the following is/are false?

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Probability for P to leave from even numbered door is 1/2

If the person left from door 2, the probability that he has entered from door 1 is (1/2)

Probability for P to leave from even numbered door is 7/12

If the person left from door 2, the probability that he has entered from door 1 is (1/4)

answer is B, C, D.

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Detailed Solution

$S = \{(1, 2), (1, 3) (1, 4), (2, 1), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 2), (4, 3)\} A) E = \{(1, 2), (1, 4) (2, 4) (3, 2) (3, 4), (4, 2)\} P (E) = \frac{6}{12} = \frac{1}{2}$
$C) P (\frac{D_{1}}{D_{2}}) = \frac{(\frac{1}{4}) (\frac{1}{3})}{(\frac{1}{4}) (\frac{1}{3}) + (\frac{1}{4}) (\frac{1}{3}) + (\frac{1}{4}) (\frac{1}{3})} = \frac{1}{3}$