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 set - theory formula

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: Analyze part a

The least - value of \(n(M\cap P)\) occurs when the two sets \(M\) and \(P\) are as disjoint as possible. Since there are no restrictions preventing them from being disjoint, the least number of students taking both courses is \(0\).

Step3: Analyze part b

The greatest value of \(n(M\cap P)\) is limited by the size of the smaller set. Since \(n(P) = 30\) and \(n(M)=90\), the maximum number of students that can be in both sets is the size of the set \(P\), so \(n(M\cap P)_{\text{max}}=30\).

Step4: Analyze part c

We know \(n(M\cup P)=n(M)+n(P)-n(M\cap P)\). 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 we want to minimize \(n(M\cup P)\), we maximize \(n(M\cap P)\). The maximum value of \(n(M\cap P)\) is \(30\). Then \(n(M\cup P)=n(M)+n(P)-n(M\cap P)=90 + 30-30=90\). Since \(n(M\cup P)+n((M\cup P)^c)=150\), then \(n((M\cup P)^c)=150 - n(M\cup P)\). Substituting \(n(M\cup P) = 90\), we get \(n((M\cup P)^c)=60\).

Answer:

a) \(0\)
b) \(30\)
c) \(60\)