Let A, B, C be pairwise independent even such that P(C) > and P(A⋂B⋂C) = 0, then P(A'⋂B'⋂C) is equal to

Engineering

Mathematics

Addition Theorem on Probability

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Basic Definitions Related to Probability

Question

Let A, B, C be pairwise independent even such that P(C) > and P(A⋂B⋂C) = 0, then P(A^'⋂B^'⋂C) is equal to

$P(A') - P(B')$

$P(A') - P(B)$

$P(A') + P(B')$

$P(A) - P(B')$

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$P (\frac{A^{c} \cap B^{c}}{C}) = \frac{P ((A^{c} \cap B^{C}) \cap C)}{P (C)}$

$= \frac{P ((1 - A) (1 - B) \cdot C)}{P (C)}$

$= \frac{P [(1 - A - B + AB) \cdot C]}{P (C)}$

$= \frac{P (C) - P (AC) - P (CB) + P (ABC)}{P (C)}$

$= \frac{P (C) - P (C) \cdot P (A) - P (C) \cdot P (B) + P (A) P (B) P (C)}{P (C)}$

$= \frac{P (C) - P (C) \cdot P (A) - P (C) \cdot P (B) + P (A \cap B \cap C)}{P (C)}$

1 – P(A) – P(B)

P(A^c) – P(B)