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"

I think its B.

Min overlap
70 + 40 = 110. Therefore min overlap is 10%. In this case the neither (brown and blue) is zero

Max overlap
100 - 70 = 30 = Neither (brown and blue)

Difference is 30% ie 0.3

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

Thankzz to all.

But i am still confused about the logic behind below concept . Dear Fluke pls elaborate :

"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"

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.

*****

Thnkxxxx a ton Fluke..U ROK MAN!

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

Hello,

I used another way to solve this one, and would like to see if it makes sense.

So, I created a table:
BE=blue eyes, NBE=not blue eyes, BH=blond hair, NBH=not blond hair. I chose 100 as total # students, and based on the information completed the third line for ALL.

............BE.......NBE......ALL
BH...............................40
NBH...................?.........60
ALL......70.........30.......100

Now, we are looking for the number that would go to ?, for two different cases (least and most).
So, my thought was that if ALL the 70 people with BE also had BH, then 0 people would have NBH, which would mean that 60 people would have NBH/NBE:


............BE.......NBE......ALL
BH.......70........................
NBH......0..........60........60
ALL......70.........30.......100

Similarly, if ALL the people with NBE would be moved to NBE/NBH, then 30 people would be at NBE/NBH:

............BE.......NBE......ALL
BH.................................
NBH..................30.......60
ALL......70.........30.......100

So, in the end 60-30 = 30. And since we have 100 people, this is 30% or 0.3.

Answer: Option

Please check the explanation as per Double Matrix Method

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Could you please explain how you reached at 30%?

When you say 30% for x is the limiting factor - do you mean that we need to pick those with no Brown hair from the set with no blue eyes i.e 30% ?

Would be a great help if you could clarify if I am correct in my understanding?

Thanks much!

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%
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Nice explanation Fluke!

Please find the solution attached.

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Let the total students be 100, Brown Hair = 40, Blue Eyes = 70
100 = 40+70 - Both + Neither
Neither = Both - 10, Neither (Minimum) = 0 --> Both = 10,
Neither (maximum) = 30 --> Both = 40
Difference = 30 --> Probability = 30/100 = 0.3

Concept for Min/Max of Neither Set A NOR Set B in a Double Set Matrix:

With no constraints given, the MINIMUM # of elements that are part of NEITHER Set A NOR Set B can = 0

the MAXIMUM # = the Smaller Value between ——(NOT A) and (NOT B)


Since there is no constraint, the MIN Probability of picking a person with NEITHER = 0/100 = 0


As for the MAX Probability:

The % of all college students that do NOT have blue eyes = 30%

The % of all college students that do NOT have brown hair = 60%

The MAX % that can have NEITHER brown eyes NOR blue eyes is limited by the Smaller of the 2 Values.

Thus, the MAX Probability = 30%

The Difference between MIN and MAX Probability = 30% - 0%

= .3

Answer -B-

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I find the easiest way is creating a double set matrix for this problem.

Let the total = 100
With the information given, we are able to determine the number that doesn't have brown hair and the number that doesn't have blue eyes. The question is what's the maximum/minimum value of the people that don't have blue eyes AND don't have brown hair.

Since we don't know the exact number of people that don't have brown hair AND have blue eyes, and we also don't know the number of people that have brown hair AND blue eyes, as well as the number of people that have brown hair AND don't have blue eyes, we can min/max these values.

The minimum of people that don't have brown hair & don't have blue eyes can be 0.
The maximum of people that don't have brown hair & don't have blue eyes can be 30.

0.3 (maximum) - 0 (minimum) = 0.3

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