Hi Bunuel,
I wanted to know whether GMAT can present such a big question. Solving it will certainly take more than 3-4 mins.

IMO : D

Consider the following Venn Diagram



No. of ppl carrying all 3 passports = 1
No. of ppl carrying none = 0


number of spies who carry Bolivian passports = two more than the number of spies who carry Argentinian passports
b+d+f+1 = a+d+e+1+2
b+f = a+e+2 --Eq (i)

number of spies who carry Bolivian passports = two less than the number of spies who carry Chilean passports
b+d+f+1 = c+e+f+1-2
b+d = c+e-2 --Eq (ii)

number of spies who carry Bolivian passports = twice the number of spies who carry both Bolivian and Chilean passports
b+d+f+1 = 2(f+1)
b+d = f+1 --Eq (iii)

number of spies who carry Bolivian passports = half the total number of spies in the room
b+d+f+1 = 1/2(a+b+c+d+e+f+1)
b+d+f+1 = a+e+c --Eq (iv)

number of spies who carry both Argentinian and Bolivian passports = two more than the number of spies who carry both Argentinian and Chilean passports
d+1 = e+1 +2
d=e+2 -- Eq (v)

number of spies who carry both Argentinian and Bolivian passports = two less than the number of spies who carry both Bolivian and Chilean passports
d+1 = f+1 -2
d = f-2 Eq (vi)

From (v) & (vi)
e+4 = f --(vii)

From (vi) & (iii)
b+f-2 = f+1
b=3

From (ii) & (iii) & (vii)
b+d = c+e-2 && b+d = f+1
f+1 = c+e-2
e+4+1 = c+e-2
c=7

From (i) & (iv)
b+f = a+e+2 && b+d+f+1 = a+e+c
a+e+2+d+1 = a+e+c
d+3=c
d=4

Sub b=3, d=4, c=7 in (ii)
e=2

From (vii)
f=6

sub b=3. c=7, d=4, e=2 , f=6 in (i)
b+f = a+e+2
9 = a+4
a=5

On solving these equations we end with
a = 5, b = 3, c = 7, d = 4, e = 2, f = 6

Prob = \(\frac{a+d+e+1}{Total}\)
= \(\frac{12}{28}\)
= \(\frac{3}{7}\)
_________________

I can not attach pic. So I am using just labels. Let

T - total

AB - A and B holder
AC - A and C holder
BC - B and C holder

a - only A holder
b - only B holder
c - only C holder.

Given:

A+2=B
C-2=B
T=2B
AB=AC+2
AB=BC-2

From these we can derive:

A+4=C
AC+4=BC

Further, if you look at the pic and differences between the numbers you will find out

a=b+2
c=b+4
c=a+2 or you can derive these by making more substitutions.

So, if you make substitution for all the numbers and express the whole table with one variable a, it will be like this.

AB=a-1
BC=a+1
AC=a-3 , in order to make it positive integer a must be more than equal 4.

Now, we can easily estimate following:

A=3a-3
B=3a-1
C=3a+1

There is a problem here. a can be any number more than equal 4 and all these number will satisfy the whole relations. I started substitution from 4.

a=4, A=9, B=11, C=13. Hence A/2B=9/22
a=5, A=12, B=14, C=16. Hence A/2B=3/7 This is in the answer choices.
a=6, A=15, B=17, C=19. Hence A/2B=15/34
a=7, A=18, B=20, C=22. Hence A/2B=9/20
a=100, A=297, B=299, C=301. Hence blaa blaa blaa

Is there max limit for a? I couldn't find

This is going to be a big explanation. It is Better to do the question with Venn diagram.
Let the number of only Argentinian spies be x
the number of only Bolivian spies be y
the number of only Chilean spies be z
the number of only Argentinian and Chilean spies be a
the number of only Argentinian and Bolivian spies be b
the number of only Chilean and Bolivian spies be c
the number of spies who carry all 3 passports is 1.

the probability that the spy selected at random will carry Argentinian passport will be

\(\frac{(x+a+b+1)}{(x+y+z+a+b+c+1)}\)

Now, moving on to the conditions

1) The number of spies who carry Bolivian passports is two more than the number of spies who carry Argentinian passports
y+b+c+1 = x+a+b+1+2
or y+c=x+a+2

2) The number of spies who carry Bolivian passports is two less than the number of spies who carry Chilean passports
y+b+c+1 =z+c+a+1-2
or y+b=z+a-2

3) The number of spies who carry Bolivian passports is twice the number of spies who carry both Bolivian and Chilean passports
y+b+c+1 = 2(c+1)
or y+b = c+1

4) The number of spies who carry Bolivian passports is half the total number of spies in the room
y+b+c+1 = 1/2(x+y+z+a+b+c+1)
or 2y+2b+2c+2 = x+y+z+a+b+c+1
or y+b+c+1= x+z+a

5) The number of spies who carry both Argentinian and Bolivian passports is two more than the number of spies who carry both Argentinian and Chilean passports
b+1 = a+1+2
or b=a+2
or a=b-2

6) The number of spies who carry both Argentinian and Bolivian passports is two less than the number of spies who carry both Bolivian and Chilean passports
b+1=c+1-2
or b=c-2
or c= b+2


Now,
putting the value of c in statement 3)

y+b= b+2+1
or y = 3

putting the value of a and c (in terms of b from statements 5 & 6 respectively) and y in statement 1)
3+b+2= x+b
or x = 5

putting value of a (in terms of b from statement 5) and y in statement 2)
3+b= z+b-4
or z = 7

putting values of x,y,z,a and c in statement 4)

3+b+b+2+1=5+7+b-2
or 2b+6 = b+10
or b = 4

putting this value of b in statement 5)

a=b-2
or a = 2

putting this value of b in statement 6)

c=b+2
or c = 6

so the probability will be

\(\frac{(x+a+b+1)}{(x+y+z+a+b+c+1)}\) = \(\frac{(5+2+4+1)}{(5+3+7+2+4+6+1)}\) = \(\frac{12}{28}\)=

\(\frac{3}{7}\)

Answer:- D

Sorry for my slow reponse. For the formula I used it is not necessecary to know the number of people who only carry Bolivian passports.
B is just the number of spies that have Bolivian passports - no matter if they have more than just this one passport.
Considering your Venn B = b + d + f + 1 = A + 2.

You don't need to solve anything to arrive at the answer. Probability of picking up a spy with Bolivian passport = 1/2
And the number of spies who hold Argentinian passport is just 2 less than that of spies who hold Bolivian passports. So the probability here will be little less than 1/2 but not a lot. 3/7 is the only option that fits. However if the options were any closer you'll have to solve for the answer.