(iii) The student has opted NSS but not NCC.
(iii) The student has opted NSS but not NCC.
Let E, F denote the events that a student has opted for NCC and NSS, respectively.
We have, P(E) =
= ![]()
P(F) =
=
and
P(E ∩ F) =
= ![]()
(i) P(E or F) = P(E ∪ F)
= P(E) + P(F) - P(E ∩ F)
= ![]()
=
=
= ![]()
Hence, P(E ∪ F) = ![]()
(ii) P(E ∪ F)’ = 1 – P(E ∪ F)
= 1 -
=
= ![]()
(iii) P(E’ ∪ F) = P(F) – P(E ∩ F) =
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