If events are mutually exclusive and , then P(A′∩B′∩C′) is equal to

Engineering

Mathematics

Conditional Probability

Question

If events are mutually exclusive and $P (A) = \frac{1}{3}, P (B) = \frac{1}{3}, P (C) = \frac{1}{4}$ , then P(A′∩B′∩C′) is equal to

$\frac{1}{4}$

$\frac{1}{8}$

$\frac{1}{3}$

$\frac{3}{4}$

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Given, $P (A) = \frac{1}{3}, P (B) = \frac{1}{3}$ and $P (C) = \frac{1}{4}$
$∴ P (A^{'}) = \frac{2}{3}, P (B^{'}) = \frac{2}{3}$ and $P (C^{'}) = \frac{3}{4}$
Now, P(A'∩B'∩C') = P(A')P(B')P(C')
(∵ A, B and C are mutually exclusive events)

$= \frac{2}{3} \times \frac{2}{3} \times \frac{3}{4} = \frac{1}{3}$

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