P(\(\overline { A } \)) = \(\frac { 1 }{ 2 } \);  P(\(\overline { A } \)) = \(1-\frac { 1 }{ 2 }=​​ \frac { 1 }{ 2 }\)
P(\(\overline { B } \)) = \(\frac { 1 }{ 3 } \);  P(\(\overline { B } \)) = 1 - \(\frac { 1 }{ 3 } \) = \(\frac { 2 }{ 3 } \)
(i) Probability that the problem is solved
= 1 - probability that the problem is not solved
\(=1-P(\overline { A } \cap \overline { B } )=1-P(\overline { A } ).P(\overline { B) } \) 
\(=1-\frac { 1 }{ 2 } \times \frac { 2 }{ 3 } =1-\frac { 1 }{ 3 } =\frac { 2 }{ 3 } \)
(ii) Probability that exactly one of them solves
\(=P(A\cap \overline { B } )+P(\overline { A } \cap B)\)
= P(A). (\(\bar { B } \)) + P(\(\overline { A } \)).P(B) 
\(=\frac { 1 }{ 2 } \times \frac { 2 }{ 3 } +\frac { 1 }{ 2 } \times \frac { 1 }{ 3 } =\frac { 2 }{ 6 } +\frac { 1 }{ 6 } =\frac { 3 }{ 6 }=\frac { 1 }{ 2 } \)

(i) Probability that both balls are red
\(=\frac { 6 }{ 25 } \times \frac { 7 }{ 25 } =\frac { 42 }{ 625 } \)
(ii) Probability that both balls are of same colour
= P(WW) + P(RR) + P(BB)
\(=\frac { 10 }{ 25 } \times \frac { 3 }{ 25 } +\frac { 6 }{ 25 } \times \frac { 7 }{ 25 } +\frac { 9 }{ 25 } \times \frac { 15 }{ 25 } \)
\(=\frac { 30+42+135 }{ 625 } =\frac { 207 }{ 625 } \)

S = {1, 2, 3,.....10}
n(S) = 10
Let B be the event of drawing a card greater than 4
B = {5, 6,.... 10}
\(P(B) =\frac { 6 }{ 10 } \)
Let A be the event of getting an even number
A = {2, 4, 6, 8, 10}
\(P(A) =\frac { 5 }{ 10 } \)
\(A\cap B\) = {6, 8, 10}  
\(P(A\cap B)=\frac { 3 }{ 10 } \)
\(P(A/B)=\frac {P(A\cap B)}{P(B)}=\frac{\frac{3}{10}}{\frac{6}{10}}=\frac{3}{6}=\frac{1}{2}\)

GM = Antilog \(\left( \frac { \sum { flogx } }{ N } \right)  \)

\(G.M=Antilog\left( \frac { 81.9092 }{ 68 } \right) \)
= Antilog(1.2045) = 16.02 tonnes

Q₁ = Size of \({ \left( \frac { N+1 }{ 4 } \right) }^{ th }\) value
= Size of (165.25)^th value = 10
Q₃ = Size of 3\({ \left( \frac { N+1 }{ 4 } \right) }^{ th }\)value
= size of (495.75)^th value = 30
Q.D = \(\frac { 1 }{ 2 } ({ Q }_{ 3 }-{ Q }_{ 1 })=\frac {30-10}{2}=\frac { 20 }{ 2 } =10\)
Co-efficient of Q.D = \(\frac { { Q }_{ 3 }-{ Q }_{ 1 } }{ { Q }_{ 3 }+{ Q }_{ 1 } } =\frac { 20 }{ 40 } =\frac { 1 }{ 2 } \) = 0.5