From the question;
The group consist of 300 students
therefore
[tex]n(\xi)\text{ = 300}[/tex]180 students take Math, 210 take English, and 145 take both
Let
Mathematics = M
English = E
Then
[tex]\begin{gathered} n(M)\text{ = 180} \\ n(E)\text{ = 210} \\ n(M\cap E)\text{ = 145} \end{gathered}[/tex]Representing the information on a venn diagram
From the venn diagram
x represent number of pupils taking maths only
Therefore
[tex]\begin{gathered} x\text{ = n(M) -n(M }\cap\text{ E)} \\ x\text{ = 180 - 145} \\ x\text{ = 35} \end{gathered}[/tex]y represents the number of pupils taking english only
Therefore,
[tex]\begin{gathered} y\text{ = n(E})\text{ - n(M }\cap E) \\ y\text{ = 210 - 145} \\ y\text{ = 65} \end{gathered}[/tex]z represents the number of students taking non of the subjects
Therefore
[tex]\begin{gathered} z\text{ = n(}\xi)\text{ - \lbrack n(M only) + n(M }\cap E)\text{ + n(E only) \rbrack} \\ z\text{ = 300 - \lbrack 35 + 145 + 65\rbrack} \\ z\text{ = 300 - 245} \\ z\text{ = 55} \end{gathered}[/tex]a. If we randomly select a student, what is the probability that the student takes Math or English, but not both?
[tex]\begin{gathered} P(\text{ maths or english but not both)} \\ =\text{ P( Maths only) + P( English only)} \\ =\text{ }\frac{n(\text{maths only)}}{\text{total students}}\text{ + }\frac{n(\text{english only)}}{total\text{ students}} \\ =\text{ }\frac{35}{300}\text{ + }\frac{65}{300} \\ =\text{ }\frac{35\text{ + 65}}{300} \\ =\text{ }\frac{100}{300} \\ =\text{ }\frac{1}{3} \end{gathered}[/tex]Therefore, the probability that the student takes Math or English, but not both is 1/3
b. If we randomly select a student, what is the probability that the student takes neither English nor Math?
[tex]\begin{gathered} P(neither\text{ maths nor english)} \\ =\text{ }\frac{no.\text{ of students taking non of the subjects}}{\text{Total students}} \\ =\text{ }\frac{55}{300} \\ =\text{ }\frac{11}{60} \end{gathered}[/tex]Therefore, the probability that the student takes neither English nor Math
is 11/60
