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
(1) The student opted for NCC or NSS
(2) The student has opted neither NCC nor NSS
(3) The student has opted NSS but not NCC

Answer 1

Hint – This type of question comes in probability. In this question, firstly we see the formula of probability. Probability is a type of ratio where we compare how many times an outcome can occur compared to all possible outcomes, i.e. probability = the no of wanted outcomes/ the no of possible outcomes. The analysis of events governed by probability is called statistics.

Complete step-by-step answer:
In this question, it is given that;
Possible no of outcomes is 60.
Let A be the event in which the selected student has opted for NCC and B be the event in which the selected student has opted for NSS.
 Number of student who have opted for NCC=30
                         $\therefore P\left( A \right) = \dfrac{{30}}{{60}} = \dfrac{1}{2}$
Also Number of student who have opted for NSS=32
                         $\therefore P\left( B \right) = \dfrac{{32}}{{60}} = \dfrac{8}{{15}}$
 Number of student who have opted for both NCC and NSS=24
                         $\therefore P\left( {{\text{A and B}}} \right) = \dfrac{{24}}{{60}} = \dfrac{2}{5}$
          (1) $P\left( {{\text{A }} \cup {\text{ B}}} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)$
                         $\therefore P\left( {A \cup B} \right) = \dfrac{1}{2} + \dfrac{8}{{15}} - \dfrac{2}{5}$
                                                   $
   = \dfrac{{15 + 16 - 12}}{{30}} \\
   = \dfrac{{19}}{{30}} \\
 $
Thus, the probability that the selected student has opted for NCC or NSS is$\dfrac{{19}}{{30}}$.

                              DEMORGAN’S LAW says:
                       $\left[ {P\left( {A' \cap B'} \right) = P\left( {A \cup B} \right)'} \right]$
              i.e. $P\left( {A'andB'} \right)$$ = P\left( {AorB} \right)'$
                         $P\left( {A \cup B} \right)' = 1 - P\left( {A \cup B} \right)$
                                              $
   = 1 - \dfrac{{19}}{{30}} \\
   = \dfrac{{11}}{{30}} \\
 $
Thus, the probability that the selected student has opted for neither NCC nor NSS is$\dfrac{{11}}{{30}}$.
(3) Number of student who have opted for NSS but not NCC is$32 - 24 = 8$
So, the probability that the selected student has opted for NSS but not NCC $ = \dfrac{8}{{60}}$$ = \dfrac{2}{{15}}$

Note – The probability of an event is a number between 0 and 1, where 0 indicates impossible event and 1 indicates certainty event. some examples of possible and impossible events, if a coin tosses no of two head comes event is $\dfrac{0}{2} = 0$ i.e. impossible event and no of event comes a head or a tell is$\dfrac{1}{2} + \dfrac{1}{2} = 1$ i.e. possible event.