If 40 percent of all students at college X have brown hair and 70 percent of all students at college X have blue eyes what is the difference between the minimum and maximum probability of picking a student from college X who has neither brown hair nor blue eyes?

Let's say there are 100 students.

40 Students BROWN hair
70 Students BLUE eyes
y Students have both brown hair and blue eyes
x Students neither Brown hair nor blue eyes

\(n(Total) = n(A)+n(B)-n(A\cap B)+n(Neither)\)
n(Total Students) = n(Brown Hair)+ n(Blue eyes) - n(Brown hair and Blue eyes) + n(Neither)
100 = 40+70-y+x
x = y-10

x will be maximum when y is maximum. What is the maximum value for
Max value for y = minimum of (40,70).
Because the number of students with brown hair is 40, the maximum number of students to have both brown hair and blue eyes can only be 40.

When y=40
x=30

Probability of selecting 30 students out of 100 students = 30/100 = 0.3

x will be minimum when y is minimum.

The minimum possible value for y=10
x = y-10
x=0

Because if "y" gets less than 10; the number of students get more than 100 which is not possible.

In this case; 0 students fit the criterion of not having either. Because all students have brown hair, blue eyes or both.

Probability becomes 0/100 = 0

Difference between two probabilities = 0.3-0 = 0.3

Ans: "B"
_________________

min prob = 0 (10% overlap b/w blue eyes and brown hair)
max prob = .7 (100% overlap b/w blue eyes and brown hair)
diff = .3

ans B

How can the number of students having neither brown hair nor blue eyes be maximized?

If the overlap between blue eyed guys and brown hair guys is increased; means the number of students having BOTH blue eyes and brown hair be increased. In terms of the set \(n(A\cap B) is increased\)

For simplicity;

Ex: Say there are 4 students(a,b,c,d);
1 student has brown hair and 2 students have blue eyes;


Stem says: 1 student has brown hair and 2 students have blue eyes;
minimum number of students who neither have blue eyes nor brown hair

Say
a-> brown hair (1 student has brown hair)
b,c-> blue eyes (2 students have blue eyes)
Number of students to have both blue eyes AND brown hair = 0
d-> doesn't have either blue eyes or brown hair. Count: 1

Maximum number of students who neither have blue eyes nor brown hair
Say
a-> brown hair & blue eyes
b-> blue eyes
Number of students to have both blue eyes AND brown hair = 1(a)
c,d-> doesn't have either blue eyes or brown hair: Count: 2

Mathematically; you can see;
To find the maximum number of students that have neither. We need to have maximum possible number of students that can have both.

Why the maximum number of students that have both can't be greater than the minimum of the two.

Maximum number of students having both = min(Students with brown hair, Students with blue eyes)
Students with brown hair = 1
Students with blue eyes = 2

Can the number of students having both be "2". NO... if out of 4 students; only 1, just ONE student has brown hair, how can there be 2 students with brown hair and blue eyes. It is clearly mentioned in the question;

Likewise;
Because the number of students with brown hair is 40, the maximum number of students to have both brown hair and blue eyes can only be 40". If only 40 students have brown hair, how can 41 students have both. It can never be any number more than 40.

Thus the maximum number of intersection is always the minimum of two counts.

Now, if 40 students have both blue eyes and brown hair and question says there are 70 students to have blue eyes. There are another 30 students who have blue eyes but no brown hair.

Total becomes 70: 40(Both brown and blue), 30(just blue).
Total students=100
Neither = 100-70=30

Coming to the find the minimum number of intersection;
if there are 40 students - brown hair
70 - blue eyes

it becomes 40+70=110. But, the total strength is 100. Thus, 10 students have both blue and brown, but 0 zero students who have neither.

*****
_________________

fluke +1 Its very nice of you to write a lengthy explanation as this one !

Answer: Option

Please check the explanation as per Double Matrix Method

_________________

You are right about it.

See if x is taken anything greater than 30% then the value in the adjacent cell to the left will be negative which is UNACCEPTABLE. Hence the maximum value of x is limited by smaller of the two adjacent summations i.e. 30% and 60%
_________________

Please find the solution attached.

_________________

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________