Question

in a survey of 150 students, 90 were taking mathematics and 30 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?

Explanation:

Step1: Recall the principle of set - theory

Let \(M\) be the set of students taking mathematics and \(P\) be the set of students taking psychology. We know \(n(M) = 90\), \(n(P)=30\) and \(n(M\cup P)+n((M\cup P)^c)=150\), and \(n(M\cup P)=n(M)+n(P)-n(M\cap P)\).

Step2: Find the least number of students taking both courses

The least number of students taking both courses occurs when the two sets \(M\) and \(P\) overlap as little as possible. Since \(n(M)+n(P)=90 + 30=120\leqslant150\), the least value of \(n(M\cap P)\) is \(0\).

Step3: Find the greatest number of students taking both courses

The greatest number of students taking both courses occurs when the smaller set is a subset of the larger set. Since \(n(P) = 30\) and \(n(M)=90\), the greatest value of \(n(M\cap P)\) is \(30\) (because the number of students taking psychology is the limiting factor).

Step4: Find the greatest number of students taking neither course

The number of students taking neither course is \(n((M\cup P)^c)=150 - n(M\cup P)\). To maximize \(n((M\cup P)^c)\), we need to minimize \(n(M\cup P)\). Since \(n(M\cup P)=n(M)+n(P)-n(M\cap P)\), and the minimum value of \(n(M\cap P)\) is \(0\), then \(n(M\cup P)_{\min}=90 + 30-0 = 120\). So \(n((M\cup P)^c)_{\max}=150 - 120=30\).

Answer:

a) 0
b) 30
c) 30