Here total number of students, n(S) = 60
Let E be the event that student opted for NCC and F be the event that the student opted for NSS.
Then n (E) = 30, n (F) = 32 and n (E ∩ F) = 24
Thus , P (E) =

n(E)

n(S)

=

30

60

=

1

2

; P (F) =

n(F)

n(S)

=

32

60

=

8

15

P (E ∩ F) =

n(E∩F)

n(S)

=

24

60

=

2

5

We know that P(E ∪ F) = P(E) + P(F) – P(E ∩ F)
(i) P(Student opted for NCC or NSS) i.e.
P(E ∪ F) =

1

2

+

8

15

−

2

5

=

15+16−12

30

=

19

30

(ii) (Student has opted neither NCC nor NSS) = P(

−

E

∩

−

F

)
= P

–––––––––

(E∪F)

= 1 - P (E ∪ F) = 1 -

19

30

=

11

30

(iii) P(Student has opted NSS but not NCC) = P (E′ ∩ F)
= P(F) − P(E ∩ F) =

8

15

−

2

5

=

8−6

15

=

2

15