Question

In a class of 60 students,30 opted for NCC, 32 opted for NSS and 24 opted for both NCC and NSS. If one of these students is selected at random, find the probability that
a) The student opted for NCC or NSS and
b) The student has opted for NSS but not NCC.

Answer 1

Hint: We start solving the problem by assigning the variables for the total numbers of students for each event. We then find the required probabilities of all the events using the number of students given in the problem. We then use the formula $P(A\cup B)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$ to find the probability that the student opted for NCC or NSS. We then use the formula \[P\left( {{A}^{'}}\cap B \right)=P\left( B \right)-P\left( A\cap B \right)\] to find the probability that the student has opted NSS but not NCC.

Complete step by step answer:
According to the problem, we are given that the total number of students in a class are 60.
Let us represent the total number of students as $n\left( S \right)=60$.
Let ‘A’ be the event of choosing the students who are opted for NCC.
According to the problem, we are given that number of students opted for NCC as 30. Let us denote it as $n\left( A \right)$. So, we get $n\left( A \right)=30$.
Now, let us find the probability for choosing the student who opted for NCC $P\left( A \right)=\dfrac{n\left( A \right)}{n\left( S \right)}=\dfrac{30}{60}=\dfrac{1}{2}$ ---(1).
According to the problem, we are given that number of students opted for NSS as 32. Let us denote it as $n\left( B \right)$. So, we get $n\left( B \right)=32$.
Now, let us find the probability for choosing the student who opted for NSS $P\left( B \right)=\dfrac{n\left( B \right)}{n\left( S \right)}=\dfrac{32}{60}=\dfrac{8}{15}$ ---(2).
According to the problem, we are given that number of students opted for both NCC and NSS as 24. Now, we get $n\left( A\cap B \right)$. So, we get $n\left( A\cap B \right)=24$.
Now, let us find the probability for choosing the student who opted for both NCC and NSS is
$P\left( A\cap B \right)=\dfrac{n\left( A\cap B \right)}{n\left( S \right)}=\dfrac{24}{60}=\dfrac{2}{5}$ ---(3).
Let us solve problem (a),
(a) The probability of student opted for NCC or NSS is represented as union of A ,B
$P(A\cup B)=P\left( A \right)+P\left( B \right)-P\left( A\cap B \right)$ ---(4).
Substitute $equation(2)\And (3)$in $equation(4)$ then we get
$P\left( A\cup B \right)=\dfrac{1}{2}+\dfrac{8}{15}-\dfrac{2}{5}=\dfrac{15+16}{30}-\dfrac{2(6)}{5(6)}=\dfrac{31-12}{30}=\dfrac{19}{30}$.
$\therefore $ $P\left( A\cup B \right)=\dfrac{19}{30}$.
(b) The probability of the student who opted for NSS but not NCC is (NSS but NCC means P(B) is excluded from intersection of A&B).
\[P\left( {{A}^{'}}\cap B \right)=P\left( B \right)-P\left( A\cap B \right)\] ---(5).
On substituting \[P\left( B \right)\And P\left( A\cap B \right)\] in \[equation(5)\] we get as follows
\[P\left( {{A}^{'}}\cap B \right)=\dfrac{8}{15}-\dfrac{2}{5}=\dfrac{8-6}{15}=\dfrac{2}{15}\].
Hence, the probability of the student who opted for NSS but not NCC is \[P\left( {{A}^{'}}\cap B \right)=\dfrac{2}{15}\].

Note: We should know that the value of the probabilities that we get should be less than or equal to 1. We can also solve this problem easily by using the Venn-diagram method, in which most of the answer can be obtained by drawing the Venn diagram itself. We should not make calculations mistakes while solving this problem.