P(A) = 0.5, P(A ∩ B) = 0.3
We have P(A U B) ≤ 1
P(A) + P(B) - P(A ∩ B) ≤ 1
0.5 + P(B) - 0.3 ≤ 1
P(B) ≤ 1 - 0.2
P(B) ≤ 0.8
Therefore, probability of B getting selected is allmost 0.8.

Total number of cards = 52; n(S) = 52
Let A be the event of getting a king card, n(A) = 4
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 4 }{ 52 } \)
Let B be the event of getting a heart card. n(B) = 13
P(B) = \(\\ \frac { n(B) }{ n(S) } =\frac { 15 }{ 52 } \)
Let C be the event of getting a red card. n(C) = 26
P(C) = \(\frac { n(C) }{ n(S) } =\frac { 26 }{ 52 } \)
P(A ∩ B) = P (getting heart king) = \(\frac{1}{52}\)
P(B ∩ C) = P (getting red and heart)) = \(\frac{13}{52}\)
P(A ∩ C) = P (getting red king) = \(\frac{2}{52}\)
P(A ∩ B ∩ C) = (getting heart, king which is red = \(\frac{1}{52}\)
Therefore, required probability is
P(A U B U C) = P(A) + P(B) + P(C) - P(A ∩ B) - P(B ∩ C) - P(C ∩ A) + P(A ∩ B ∩ C)
=\(\frac { 4 }{ 52 } +\frac { 13 }{ 52 } +\frac { 26 }{ 52 } -\frac { 1 }{ 52 } -\frac { 13 }{ 52 } -\frac { 2 }{ 52 } +\frac { 1 }{ 52 } =\frac { 28 }{ 52 } =\frac { 7 }{ 13 } \).

Total number of cards = 52
Number of king cards = 4
Probability of drawing a king card = \(\frac{4}{52}\)
Number of queen cards = 4
Probability of drawing a queen card = \(\frac{4}{52}\)
Both the events of drawing a king and a queen are mutually exclusive
⇒ P(AUB) = P(A) + P(B)
Therefore, probability of drawing either a king or a queen = \(\frac { 4 }{ 52 } +\frac { 4 }{ 52 } =\frac { 2 }{ 13 } \).

P(A) = 0.37, P(B) = 0.42, P(A∩B) = 0.09
P(AUB) = P(A) + P(B) - P(A∩B)
P(AUB) = 0.37 + 0.42 - 0.09 = 0.7

Mean \(\bar { x } \) = 25.6, Coefficient of variation, C.V. = 18.75
Coefficient of variation, C.V. = \(\frac { \sigma }{ \bar { x } } \) x 100%
18.75 = \(\frac { \sigma }{ 25.6 } \) x 100; σ = 4.8

CV =\(\frac { \sigma }{ \bar { x } } \) x 100 ⇒ \(\bar { x } =\frac { \sigma }{ CV } \) x 100
\(\bar { x } =\frac { 15.6 }{ 6.9 } \) x 100  = 22.6

Mean \(\bar { x } \)  = \(\frac { 16+13+17+21+18 }{ 5 } =\frac { 85 }{ 5 } \)  = 17

σ =\(\sqrt { \frac { \Sigma d^{ 2 } }{ n } } =\sqrt { \frac { 34 }{ 5 } } =\sqrt { 638 } \)
σ = 2.61
Co-efficient of variation
CV = \(\frac { \sigma }{ \bar { x } } \) x 100 = \(\frac { 2.61 }{ 17 } \) x 100
= 15.35%

Σ(x -10)² = 79 = Σx²- 20x + 100 = 79
= Σx²- 20Σx + 100 x 9 = 79
= Σx²- 20 x 99 + 900 = 79
Σx² = 79 + 1980 - 900 = 1159
Σ(x - \(\bar { x } \))² = Σ(x - 11)² = Σ(x² - 22x + 121)
= Σx² - 22Σx + 121 x 9
= 1159 - 22 x 99 + 1089 = 70
∴ Σx² = 1159, Σ(x - \(\bar { x } \))² = 70

Total number of students n(S) = 50.
Let A and B be the events of students opted for NCC and NSS respectively.
n(A) = 28, n(B) = 30, n(A⋂B) = 18
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 28 }{ 50 } \)
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 30 }{ 50 } \)
P(A∩B) = \(\frac { n(A\cap B) }{ n(S) } =\frac { 18 }{ 50 } \)
(i) Probability of the students opted for NCC but not NSS
P(A ∩ \(\bar { B } \)) = P(A) - P(A ∩ B) = \(\frac { 28 }{ 50 } -\frac { 18 }{ 50 } =\frac { 1 }{ 5 } \)
(ii) Probability of the students opted for NSS but not NCC.
P(A ∩ \(\bar { B } \)) = P(B) - P(A ∩ B) = \(\frac { 30 }{ 50 } -\frac { 18 }{ 50 } =\frac { 6 }{ 25 } \)
(iii) Probability of the students opted for exactly one of them
= P[(A ∩ \(\bar { B } \)) U (\(\bar { A } \) ∩ B)]
= P(A ∩ \(\bar { B } \)) + P(\(\bar { A } \) ∩ B) =\(\frac { 1 }{ 5 } +\frac { 6 }{ 25 } =\frac { 11 }{ 25 } \)
(Note that (A ∩ \(\bar { B } \)),(\(\bar { A } \) ∩ B) are mutually exclusive events)

First we find the coefficient of variation for guavas and oranges separately.
Number of guavas, n=7

Mean \(\bar { { x }_{ 1 } } \)=\(\frac { 30 }{ 7 } \)=4.29
Standard deviation σ₁=\(\sqrt { \frac { \Sigma { x }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma x }_{ i } }{ n } \right) ^{ 2 } } \)
σ₁=\(\sqrt { \frac { 136 }{ 7 } -\left( \frac { 30 }{ 7 } \right) ^{ 2 } } =\sqrt { 19.43-18.40 } \) ≃ 1.01
Coefficient of variation for guavas
C.V₁=\(\frac { { \sigma }_{ 1 } }{ \bar { { x }_{ 1 } } } \times 100\)% = \(\frac { 1.01 }{ 4.29 } \) x 100% =23.54%

 Number of oranges n=7
Mean \(\bar { { x }_{ 2 } } \)=\(\frac { 30 }{ 7 } \)=4.29
Standard deviation σ₁=\(\sqrt { \frac { \Sigma { x }_{ i }^{ 2 } }{ n } -\left( \frac { { \Sigma x }_{ i } }{ n } \right) ^{ 2 } } \)
σ₂=\(\\ \sqrt { \frac { 184 }{ 7 } -\left( \frac { 30 }{ 7 } \right) ^{ 2 } } =\sqrt { 26.29-18.40 } \)=2.81
Coefficient of variation for oranges
C.V₁=\(\frac { { \sigma }_{ 2 } }{ \bar { { x }_{ 2 } } } \) x 100% = \(\frac { 2.81 }{ 4.29 } \) x 100% = 65.50%
CV₁=23.54%, CV₂=65.50%. Since, C.V₁ < C.V₂,  we can conclude that the consumption of guavas is more consistent than oranges.