Given:
The total number of students, N=160.
The number of students taking sequence course, n(S)=76.
The number of students taking analysis course, n(A)=85.
The number of students taking sequence and analysis courses , n(S∩A)=33.
The number of students taking analysis and probability courses, n(A∩P)=35.
The number of students taking sequence, analysis and probability courses, n(S∩A∩P)=15.
The number of students taking only sequence course, n(s)=32.
Since each student takes atleast one math course, the total number of students is,
[tex]n(\text{S}\cup\text{A}\cup P)=N=160[/tex]The venn diagram formula can be written as,
[tex]n(\text{S}\cup\text{A}\cup P)=n(S)+n(A)+n(P)-n(S\cap A)-n(A\cap P)-n(S\cap P)+n(S\cap A\cap P)[/tex]We use Venn diagram to solve the question. Let S represenst students taking sequence course, A represenst students taking analysis course and P represenst students taking probability course.
The students taking only analysis and sequence courses is,
[tex]n\mleft(S\cap A\mright)-n\mleft(S\cap A\cap P\mright?)=33-15=18[/tex]The students taking only analysis and probability courses is,
[tex]n\mleft(A\cap P\mright)-n\mleft(S\cap A\cap P\mright)=35-15=20[/tex](b)
The students taking only sequence and probability lessons without analysis is,
[tex]\begin{gathered} n(S\cup P)=n(S)-18-15-32 \\ =76-18-15-32 \\ =11 \end{gathered}[/tex]Hence, the number of students taking only sequence and probability lessons without analysis is 11.
(a)
Using the Venn diagram, the number of students taking only probability course is,
n(p)=160-32-18-11-15-32-20=32
Now, the number of students taking a probability course is,
n(P)=11+15+20+32=78.
Therefore, the number of students taking a probability course is 78.
(c)
The students that take only one course from the venn diagram is
N=32+32+32=96
Therefore, the students that take only one course is 96.
