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Let E^c denote the complement of an even...

Let `E^c` denote the complement of an event `E.` Let `E,F,G` be pairwise independent events with `P(G) gt 0` and `P(E nn F nn G)=0` Then `P(E^c nn F^c nn G)` equals (A) `P(E^c)+P(F^c)` (B) `P(E^c)-P(F^c)` (C) `P(E^c)-P(F)` (D) `P(E)-P(F^c)`

A

`P(E^(c ))+P(F^(c ))`

B

`P(E^(c ))-P(F^(c ))`

C

`P(E^(c ))-P(F)`

D

`P(E)-P(F^(c ))`

Text Solution

Verified by Experts

The correct Answer is:
C
`P((E^(c) cap F^(c))/(G))=(P(E^(c)cap F^(c)capG))/(P(G))=(P(G)-P(EcapG)-P(GcapF))/(P(G))=(P(G)[1-P(E)-P(F)])/(P(G))" "[ :' P(G) ne 0]`
`=1-P(E)-P(F)=P(E^(c)-P(F)`
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