Question

in a survey of 150 students, 80 were taking mathematics and 10 were taking psychology.
a) what is the least number of students who could have been taking both courses?
b) what is the greatest number of students who could have been taking both courses?
c) what is the greatest number of students who could have been taking neither course?
a) the least number of students who could have been taking both courses is

b) the greatest number of students who could have been taking both courses is

c) the greatest number of students who could have been taking neither course is

Explanation:

Step1: Recall the formula for the union of two sets

The formula for \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\), where \(n(A)\) is the number of elements in set \(A\), \(n(B)\) is the number of elements in set \(B\), \(n(A\cup B)\) is the number of elements in the union of \(A\) and \(B\), and \(n(A\cap B)\) is the number of elements in the intersection of \(A\) and \(B\). Also, \(n(A\cup B)\leq N\) (where \(N\) is the total number of students) and \(n(A\cap B)\geq0\).

Step2: Solve for the least - value of \(n(A\cap B)\) (part a)

We know that \(n(A) = 80\), \(n(B)=10\), and \(N = 150\).
From \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\), we can rewrite it as \(n(A\cap B)=n(A)+n(B)-n(A\cup B)\).
Since \(n(A\cup B)\leq N\), to minimize \(n(A\cap B)\), we maximize \(n(A\cup B)\).
The maximum value of \(n(A\cup B)\) is \(150\).
So \(n(A\cap B)=80 + 10-150=-60\). But \(n(A\cap B)\geq0\), so the least value of \(n(A\cap B)\) is \(0\).

Step3: Solve for the greatest - value of \(n(A\cap B)\) (part b)

Since \(n(A\cap B)\leq\min(n(A),n(B))\), and \(n(A) = 80\), \(n(B)=10\).
The greatest value of \(n(A\cap B)\) is \(\min(80,10)=10\).

Step4: Solve for the greatest - value of \(n(\text{neither})\) (part c)

We know that \(n(\text{neither})=N - n(A\cup B)\).
From \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\), to maximize \(n(\text{neither})\), we minimize \(n(A\cup B)\).
The minimum value of \(n(A\cup B)\) occurs when \(n(A\cap B)\) is maximized.
Since \(n(A\cup B)=n(A)+n(B)-n(A\cap B)\), when \(n(A\cap B) = 10\) (the maximum value of \(n(A\cap B)\)), \(n(A\cup B)=80 + 10-10=80\).
Then \(n(\text{neither})=150 - 80=70\).

Answer:

a) \(0\)
b) \(10\)
c) \(70\)