n(S) = 5C₂
(i) Let A be the event that the committee has no woman
n(A) = 3C₂
P(A) = \(\frac { 3{ C }_{ 2 } }{ 5{ C }_{ 2 } } =\frac { 3C_1 }{ \frac { 5\times 4 }{ 2\times 1 } } = \frac { 3 }{ 10 } \)
(ii) Let B be the event that the committee will have one man
n(B) = 3C₁ x 2C₁= 3 x 2 = 6
P(B) =\(\frac { 6 }{ 5{ C }_{ 2 } } =\frac {6 }{\frac { 5\times 4 }{ 2\times 1 }} =\frac { 6 }{ 10 }=\frac { 3 }{ 5 } \)
(iii) Let C be the event that the committee will have no man
n(C) = 2C₂ = 1
P(C) =\(\frac {1}{ 5{ C }_{ 2 } } =\frac { 1 }{ 10 } \)

Let A, B be the events of getting a black ball in the first and second draw
Probability of drawing a black ball in the first attempt is
P(A) = \(\frac{3}{5+3}=\frac{3}{8}\)
Probability of drawing the second black ball given that the first ball drawn is black
P(B/A) = \(\frac{2}{5+2}=\frac{2}{7}\)
\(\therefore\) The probability that both balls drawn are black is given by
\(P\left( A\cap B \right) =P(A)P(B/A)=\frac { 3 }{ 8 } \times \frac { 2 }{ 7 } =\frac { 3 }{ 28 } \)

We know that the sample space is S = {1,2,3,4,5,6}
Now, A = {3,6} ; B = { 2,4,6} then (A∩B) = {6}
P(A) = \(\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)
P(B) = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \) and P(A∩B) = \(\frac{1}{6}\)
Clearly P(A∩B) = P(A) P(B)
Hence A and B are independent events.

S = {(1, 1) (1, 2) (1,3), ..... (6, 6)}
(2, 1), (2, 2) ...... (2, 6)
(6, 1), (6, 2)  ..... (6, 6)}
n(S) = 36
Let B be the event that sum of numbers appearing is 6
B = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
\(P(B)=\frac{5}{36}\)
Let A be the event that 4 has appeared atleast once
A = {(2, 4), (4, 2)}
\(A\cap B\)  = {(2, 4),(4, 2)}
\(P(A\cap B)=\frac{2}{36}\)
\(P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { 2/36 }{ 5/36 } =\frac { 2 }{ 5 } \)

Arranging in ascending order
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22
n = 11
 Q₁ = size of \((\frac { { n+1} }{ 4 } )^{ th }\)  value
= size of \({ \left( \frac { 11+1 }{ 4 } \right) }^{ th }\) value
= size of 3^rd value = 6
Q₃ = Size of \({ \left( \frac { 3(n+1) }{ 4 } \right) }^{ th }\) value
= size of \({ \left( 3\left( \frac { 11+1 }{ 4 } \right) \right) }^{ th }\) value
=size of 3(3)^th value
= size of 9^th value = 18