\(\sum { log } \) = 4.3167
Geometric Mean = Antilog\(\left( \frac { \sum { log\quad x } }{ N } \right) =Antilog\left( \frac { 4.3167 }{ 4 } \right) \)
= Antilog (1.0791)
GM = 11.99

(i) S = {1, 2, 3, 4, 5, 6}
n(S) = 6
(i) Let A be the event of getting a prime number
A = {2, 3, 5}
n(A) = 3
\(P(A)=\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
(ii) Let B be the event that the number is greater than or equal to 3.
B = {3, 4, 5, 6}    
n(B) = 4
\(P(B) =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)

(i) Here the distance is constant. Hence harmonic mean is appropriate.
\(HM=\frac { n }{ \frac { 1 }{ a } +\frac { 1 }{ b } } \)
\(=\frac { 2 }{ \frac { 1 }{ 40 } +\frac { 1 }{ 30 } } =\frac { 2 }{ \frac { 7 }{ 120 } } \)
\(\frac { 2\times 120 }{ 7 } \) = 34.3 km per litre
(ii) Here the quantity of petrol consumed is fixed i.e 2 litres. Here the arithmetic mean will give the correct answer.
\(\overset { - }{ X } =\frac{Total \ distance \ Covered}{Total \ Petrol \ Consumed}\)
\(\overset { - }{ X } =\frac { 40\times 2+30\times 2 }{ 4 } =35\)
\(\therefore\)  Average speed = 35 km per litre.

If the problem is given to a layman he is most likely to compute the arithmetic mean of two speeds 
i.e., \(\overset { - }{ X } =\frac { 30km+20km }{ 2 } =25kmph\)
But this is not the correct average.
So harmonic mean would be mean suitable in this situation. Harmonic Mean of 30 and 20 is
\(HM=\frac { 2 }{ \left( \frac { 1 }{ 10 } \right) +\left( \frac { 1 }{ 30 } \right) } =\frac { 2 }{ \left( \frac { 5 }{ 60 } \right) } =\frac { 2\left( 60 \right) }{ 5 } \)
= 24 kmph

GM = Anti \(\log { \left( \frac { \sum { flog \ m } }{ N } \right) } \)
= Anti \(\log { \left( \frac { 67.1394 }{ 52 } \right) } \)
= Anti log(1.2911)
GM = 19.54

GM = Anti \(\log { \left( \frac { \sum { logX } }{ n } \right) } ;\) where n = 10
GM = Anti log \(\left( \frac { 17.6373 }{ 10 } \right) \)
= Anti log(1.7637)
GM = 58.03

\(n(S)=52{ C }_{ 2 }=\frac { 52\times 51 }{ 2\times 1 } =1326\)
Let A be the event that one king and one queen card is drawn
n(A) = 4C₁ \(\times\) 4C₁ = 16
\(P(A)=\frac { 16 }{ 1326 } =0.012\)

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)}
\(P(A\cap B)=\frac{2}{36}\)
\(P(A/B)=\frac { P(A\cap B) }{ P(B) } =\frac { 2/36 }{ 5/36 } =\frac { 2 }{ 5 } \)

S {(G₁, G₂), (G₁, B₂), (B₁, G₂), (B₁, B₂)}
n(S) = 4
Let A be the event that both are girls
A= {(G₁ G₂)}
n(A) = 1
\(P(A)=\frac{1}{4}\)
Let B to the event that atleast one of the children is a girl.
B = {(G₁ G₂), (G₁ B₁), (G₂ B₂)}
\(P(B)=\frac{3}{4}\)
\( A\cap B=\{ ({ G }_{ 1 }{ G }_{ 2 })\}, P(A\cap B)=\frac{1}{4}\)
\(P(A / B)=\frac { P(A\cap B) }{ P(B) } =\frac{\frac{1}{4}}{\frac{3}{4}}=\frac { 1 }{ 3 } \)

Let A = {First ball drawn is green}
B = {Second ball drawn is green}
\(P(A)=\frac { n(A) }{ n(S) } =\frac { 6 }{ 10 } \)   [\(\because\) Total number of balls = 4+6 = 10]
Let C = {getting green ball after taking out first green ball}
\(P(C)=\frac { 5{ C }_{ 1 } }{ 9{ C }_{ 1 } } =\frac { 5 }{ 9 } \)   [\(\because\) First ball is not replaced]
\(\therefore\) P(getting green ball) = P(A).P(C) = \(\frac { 6 }{ 10 } \times \frac { 5 }{ 9 } \)
\(\therefore P(A\cap B)=\frac { 1 }{ 3 } \)
\(P(B/A)=\frac { P(A\cap B) }{ P(A) } =\frac { \frac { 1 }{ 3 } }{ \frac { 6 }{ 10 } } \)
\(P(B/A)=\frac { 1 }{ 3 } \times \frac { 10 }{ 6 } =\frac { 5 }{ 9 } \)

Total Number of bolts produced = 1000 + 2000 + 3000 = 6000
\(P({ A }_{ 1 })=\frac { 1000 }{ 6000 } =\frac { 1 }{ 6 } \)
\(P({ A }_{ 2 })=\frac { 2000 }{ 6000 } =\frac { 1 }{ 3 } \)
\(P({ A }_{ 3 })=\frac { 3000 }{ 6000 } =\frac { 1 }{ 2 } \)
Let B be the event of selecting defective bolts.
\(\therefore \) P(B/A₁) = 1% = \(\frac { 1 }{ 100 } \) = 0.01
P(B/A₂) = 1.5% = 0.015
and (P(B/A₃) = 2% = 0.02
\(\therefore P(A_{ 1 }/B)=\frac { P({ A }_{ 1 }).P\left( B/{ A }_{ 1 } \right) }{ P({ A }_{ 1 }).P(B/{ A }_{ 1 })+P({ A }_{ 2 }).P\left( B/{ A }_{ 2 } \right) +P({ A }_{ 3 }).P\left( B/{ A }_{ 3 } \right) } \)
\(=\frac { \frac { 1 }{ 6 } \times 0.01 }{ \frac { 1 }{ 6 } \times 0.01+\frac { 1 }{ 3 } \times 0.015+\frac { 1 }{ 2 } \times 0.02 } =\frac { 1 }{ 600 } \times 60=\frac { 1 }{ 10 } \)
\(\therefore P(A_{ 1 }/B)=0.1\)

Here the sample space of the experiment is
S = {HHH, HHT, HTH, HTT, THH, TTH, THT, TTT}
A = {Three heads or Three tails}
= {HHH, TTT}
B = {at least two heads}
= {HHH, HHT, HTH, THH} and
C = {at most two heads} = {HHT, HTH, HTT, THH, TTH, THT, TTT}
Also (A∩B) = {HHH}; (A∩C) = {TTT} and (B∩C) ={HHT, HTH, THH}
∴ P(A) = \(\frac { 2 }{ 8 } =\frac { 1 }{ 4 } \); P(B) =\(\frac{1}{2}\); P(C) = \(\frac{7}{8}\) and
P(A∩B)= \(\frac{1}{8}\), P(A∩C)=\(\frac{1}{8}\), P(B∩C)=\(\frac{3}{8}\)
Also P(A). P(B)=\(\frac { 1 }{ 4 } .\frac { 1 }{ 2 } =\frac { 1 }{ 8 } \)
P(A). P(C) =\(\frac { 1 }{ 4 } .\frac { 7 }{ 8 } =\frac { 7 }{ 32 } \)
and P(B). P(C) =\(\frac { 1 }{ 2 } .\frac { 7 }{ 8 } =\frac { 7 }{ 16 } \)
Thus, P(A∩B) = P(A). P(B)
P(A∩C) ≠ P(A) P(C) and
P(B∩C) ≠ P(B). P(C)
Hence, the events (A and B) are independent, and the events (A and C) and (B and C) are dependent.

Let the events E₁, E₂ and A be defined as
E₁ - event that male subscriber is under 30
E₂ - event that male subscriber is above 30
A is event that subscriber is male
P(E₁) = 0.80 and P(E₂) = 1 - 0.8 = 0.2
P(B/E₁) = 0.20, P(B/E₂) = 0.30
\(P({ E }_{ 1 }/B)=\frac { P({ E }_{ 1 }).P\left( \frac { B }{ { E }_{ 1 } } \right) }{ P({ E }_{ 1 }).P\left( \frac { B }{ { E }_{ 1 } } \right) +P({ E }_{ 2 }).P\left( \frac { B }{ { E }_{ 2 } } \right) } \)
\(=\frac { 0.8\times 0.2 }{ 0.8\times 0.2+0.2\times 0.3 } \)
\(=\frac { 0.16 }{ 0.16+0.06 } =\frac { 16 }{ 22 } =0.727\)

Let the events E₁, E₂, E₃ and A be defined as
E₁ - Cars in the service station A
E₂ - Cars in the service station B
E₃ - Cars in the service station C
A is Event of failing to clean the glass
\(P({ E }_{ 1 })=\frac { 50 }{ 100 } ,P({ E }_{ 2 })=\frac { 30 }{ 100 } ,P({ E }_{ 3 })=\frac { 20 }{ 100 } \)
\(P(A/E_{ 1 })=\frac { 5 }{ 100 } ,P(A/{ E }_{ 2 })=\frac { 7 }{ 100 } ,P(A/{ E }_{ 3 })=\frac { 3 }{ 100 } \)
Probability that the glass is not cleaned = P(A)
= P(E₁).P(A/E₁)+P(E₂).P(A/E₂)+P(E₃).P(A/E₃)
\(=\frac { 50 }{ 100 } \times \frac { 5 }{ 100 } +\frac { 30 }{ 100 } \times \frac { 7 }{ 100 } +\frac { 20 }{ 100 } \times \frac { 3 }{ 100 } \)
\(=\frac { 250+210+60 }{ 10,000 } \)
\(=\frac { 520 }{ 10,000 } \)
= 0.052
Probability that glass is cleaned
= P(A') = 1 - P(A)
= 1 - 0.052 = 0.948