If P(E)=1/3, P(F)=1/4, and P(E n F)=1/6 find?

Question

1-P(E U F)
2-P(E' n F)

note: n=intersection
P=probability
U=union

plz help me understand

Answer 1

1- P(E U F)=P(E) +P(F)-P(E n F)
=1/3 +1/4 -1/6=5/12
2- P(E'nF)=P(F)-P(EnF)
=1/4-1/6=1/12

Answer 2

this looks absolutely weird to me. i've never seen probability and sets combined in a single problem... but i'll try to help you based on whatever i know and some random wild guesses...

draw a venn diagram:
P(E) = 1/3
P(F) = 1/4
P(E n F) = 1/6

{ [ 1/6 ( 1/6 ] 1/12 ) 7/12 }

[ ] : E
( ) : F
{ } : universal set

i hope you understand what i'm trying to say in my one-line venn diagram. this is what i interpret the question as:

P(E) = 1/6 + 1/6 = 1/3
P(F) = 1/12 + 1/6 = 1/4
P(universal set) = 1

so the first 1/6 you see is part of E [i.e. (E n F')]. the second 1/6 is the area of intersection between E and F [i.e. (E n F)]. the 1/12 is part of F [i.e. (F n E')]. and the 7/12 is (E U F)', because [1 - (1/6 + 1/6 + 1/12) = 7/12].

so once you understand this, the rest should be pretty straightforward.

1. P(E U F) = 1/6 + 1/6 + 1/12 = 5/12

2. P(E' n F) = 1/12

Answer 3

assume this is the area of squares.
you have one square with area 1
inside you have square E with area 1/3 and F with area 1/4.
E and F are overlapping with area of 1/16.
1. P(E U F) = area of E + area of F - the overlapping area (since it is counted once and not twice) = 1/3 + 1/4 - 1/16 = ? ;-)
1. P(E' n F) = area outside E that is overlapping F = area of F - overlapping area with E (since this is the area of F that is not overlapping E) = 1/4 - 1/16 = ? ;-)

I'll let you do the rest.