There are two ways to do it;
(1) By Venn diagram
(2) By formula
I'll do it both ways:
(1) By Venn diagram:
Let the probabilities of the four regions in the Venn
diagram be a,b,c, and d:
Then if we understand what
"one set given another" means, we have:
P(E|F) = b/(b+c) = 3/4
P(F|E) = b/(a+b) = 1/2
P(EUF) = (a+b+c)/(a+b+c+d) = 2/3
P(U) = a+b+c+d = 1
Cross multiplying and simplifying on the first three,
we have the system:
Subtracting the 3rd equation from the 4th,
d = 1-2d
3d = 1
d = 1/3
Since b=3c, c=b/3
and we have b=a
Substitute those and d = 1/3 in
a+b+c=2d
b+b+b/3 = 2(1/3)
2b+b/3 = 2/3
Multiply through by 3
6b+b = 2
7b = 2
b = 2/7
a = b = 2/7
c = b/3 = (2/7)/3 = 2/21
So Pr(E) = a+b = 2/7+2/7 = 4/7
and Pr(F) = b+c = 2/7+2/21 = 8/21
--------------------------------
By formulas, which requires no understanding:
Pr(E|F) = Pr(E&F)/Pr(F) = 3/4
Pr(F|E) = Pr(F&E)/Pr(E) = 1/2
Pr(EUF) = Pr(E)+Pr(F)-Pr(E&F) = 2/3
From the first: 3Pr(F) = 4Pr(E&F)
Pr(F) = 4Pr(E&F)/3
From the second: Pr(E) = 2Pr(F&E)
Substituting in:
Pr(E)+Pr(F)-Pr(E&F) = 2/3
2Pr(F&E)+4Pr(E&F)/3-Pr(E&F) = 2/3
Since F&E = E&F
2Pr(E&F)+4Pr(E&F)/3-Pr(E&F) = 2/3
Pr(E&F)+4Pr(E&F)/3 = 2/3
Multiply through by 3
3Pr(E&F)+4Pr(E&F) = 2
7Pr(E&F) = 2
Pr(E&F) = 2/7
Substituting in
Pr(E) = 2Pr(F&E)
Pr(E) = 2Pr(E&F)
Pr(E) = 2(2/7)
Pr(E) = 4/7
Pr(F) = 4Pr(E&F)/3
Pr(F) = 4(2/7)/3
Pr(F) = (8/7)/3
Pr(F) = 8/21
Edwin
