Its all about Complements.....

Probability Level 4

Let $E^C$ denote the complement of an event E . Let E,F,G be pairwise independent events with P(G)>0 and $P( E ∩ F ∩ G) = 0$ . Then $P( E^C ∩ F^C | G)$ equals

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P (E) - P (F^C)

P (E^C) - P (F)

P (E^C) - P (F^C)

P (E^C) + P (F^C)

1 solution

Harshvardhan Mehta
Sep 29, 2014

$P( {E^C ∩ F^C} | {G} ) = \frac{P(E^C ∩ F^C ∩ G)}{P(G)}$ $= \frac{P(G) - P(E ∩ G) - P(G ∩ F)}{P(G)}$ $= \frac{P(G) - (1 - P(E) - P(F))}{P(G)} [P(G) \neq 0]$ $= 1 - P(E) - P(F)$ $= P(E^C) - P(F)$

I think you are asking for conditional probability, which is written as $P ( A | B )$ instead of $P ( \frac{A}{B} )$ . Can you confirm if this is what you intended?

Calvin Lin Staff - 6 years, 7 months ago

@Calvin Lin yes this is what i meant. and yes thanx for correcting it.

Harshvardhan Mehta - 6 years, 7 months ago

No problem :) Can you update your solution to reflect this?

Calvin Lin Staff - 6 years, 7 months ago

Its all about Complements.....

1 solution

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