Bunuel wrote:
In a roomful of spies, some spies carry Argentinian passports, some carry Bolivian passports, and some carry Chilean passports. Every spy carries at least one passport, no spy carries more than one passport from a given country, and only one spy carries passports from all three countries. The number of spies who carry Bolivian passports is two more than the number of spies who carry Argentinian passports, two less than the number of spies who carry Chilean passports, twice the number of spies who carry both Bolivian and Chilean passports, and half the total number of spies in the room. 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 and two less than the number of spies who carry both Bolivian and Chilean passports. If a spy is selected at random, what is the probability that he will carry an Argentinian passport?
(A) 2/9
(B) 13/57
(C) 6/23
(D) 3/7
(E) 1/2
Kudos for a correct solution.
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 = 3putting 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 = 7putting 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 = 4putting this value of b in statement 5)
a=b-2
or
a = 2putting this value of b in statement 6)
c=b+2
or
c = 6so 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