Question

In a class of 175 students, the following data shows the number of students opting one or more subjects. Mathematics 100; Physics 70; Chemistry 40; Mathematics and Physics 30; Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and Chemistry 18. How many students have offered mathematics alone?
A. 35
B. 48
C. 60
D. 22

Answer 1

We first explain the given numerical forms of the $n\left( A \right),n\left( B \right),n\left( A\cap B \right),n\left( A\cap B\cap C \right)$ where sets A, B and C are for students opting for Mathematics, Physics and Chemistry respectively. We use $n\left( A\cap {{B}^{c}}\cap {{C}^{c}} \right)=n\left( A \right)-n\left( A\cap B \right)-n\left( A\cap C \right)+n\left( A\cap B\cap C \right)$ to express the students’ option for Mathematics alone. We put the values to find the solution.

Complete answer:
The given problem is the problem of set inclusion.
We assume three sets A, B and C for students opting for Mathematics, Physics and Chemistry respectively. S denotes the total number of students.
In total there are 175 students out of which 100 in Mathematics; 70 in Physics; 40 in Chemistry.
So, $n\left( S \right)=175$, $n\left( A \right)=100$, $n\left( B \right)=70$ and $n\left( C \right)=40$.
We have students in Mathematics and Physics 30; Mathematics and Chemistry 28; Physics and Chemistry 23; Mathematics, Physics and Chemistry 18.
So, $n\left( A\cap C \right)=28$, $n\left( B\cap C \right)=23$, $n\left( A\cap B \right)=30$ and $n\left( A\cap B\cap C \right)=18$.
We have to find the number of students that have offered mathematics alone. The set can be denoted by $n\left( A\cap {{B}^{c}}\cap {{C}^{c}} \right)$. We are taking complementary events.
From set inclusion $n\left( A\cap {{B}^{c}}\cap {{C}^{c}} \right)=n\left( A \right)-n\left( A\cap B \right)-n\left( A\cap C \right)+n\left( A\cap B\cap C \right)$.
Putting the values, we get $n\left( A\cap {{B}^{c}}\cap {{C}^{c}} \right)=100-30-28+18=60$.
And hence the correct answer is option C.

Note:
We need to remember that the relation between numerical values and their probabilities is similar for all the given $n\left( A \right),n\left( B \right),n\left( A\cup B \right),n\left( A\cap B \right)$. That’s why we didn’t use the concept of number of points in a set and instead we directly used the numerical form to find the solution. To find individual probabilities, we divide them with $n\left( S \right)$.