Given, a group consists 12 persons.
\(\therefore \)  Total number of outcomes = 12
(i) Given, number of extremely patient persons = 3
\(\therefore \)  Number of favourable outcomes = 3
\(\therefore \)  P (extremely patient) = \(\frac{3}{12} = \frac{1}{4}\)
(ii) Given, number of extremely honest persons = 6 and number of extremely kind  persons
= 12 - 6 - 3 = 3
\(\therefore \)  Number of favourable outcomes = Number of extremely kind persons + Number of extremely honest persons = 6 + 3 = 9
\(\therefore \)  P (extremely kind or honest) = \(\frac{9}{12}=\frac{3}{4}\)

Remaining cards of spade=13-3=10
So, favourable outcomes=10,
i.e. n(E₄)=10
∴ P(getting a spade)​​​​​​​\(=\frac { 10 }{ 49 } \)

When two coins are tossed simultaneously, all possible outcomes are {HH, HT, TH, TT}.
Total number of sample space is n(S)=4
(i) Let E₁=Event of getting no head={TT}
n(E₁)=1
∴ P (getting no head)\(=\frac { n(E_{ 1 }) }{ n(S) } =\frac { 1 }{ 4 } \)
(ii) Let E₂=Event of getting at most one tail
={HT, TH, HH}
n(E₂)=3
∴ P (getting atmost one tail)\(=\frac { n(E_{ 2 }) }{ n(S) } =\frac { 3 }{ 4 } \)
(iii) Let E₂=Event of getting one tail={HT, TH}
n(E₃)=2
∴ P (getting one tail)\(=\frac { n(E_{ 3 }) }{ n(S) } =\frac { 2 }{ 3 } =\frac { 1 }{ 2 }\)
(d) Let E₄=Event of getting one head and one tail
={HT, TH}
n(E₄)=2
∴ P (getting one head and one tail)\(=\frac { n(E_{ 4 }) }{ n(S) } =\frac { 2 }{ 4 } =\frac { 1 }{ 2 }\)

\({2\over3},{8\over23}\)

When two dice are thrown simultaneously, then sample space contain ( 6 X6=36) outcomes
Sum 2 i.e., [(1,1),(1,1)] = two outcomes
P( Sum 2) = 2 / 36 = 1 / 18
Sum 3 i.e., (1,2), (1,2), (2,1), (2,1) =4 outcomes.
P( Sum 3) = 4 / 36 = 1 / 9
Sum 4 i.e., [(1,3), (1,3), (2,2), (2,2),(3,1), (3,1)] =6 outcomes
P( Sum 4) = 6 / 36 = 1 / 6
Sum 5 i.e., [(2,3), (2,3), (3,2), (3,2),(4,1), (4,1)] =6 outcomes
P( Sum 5) = 6 / 36 = 1 / 6
Sum 6 i.e., [(3,3), (3,3), (4,2), (4,2),(5,1), (5,1)] =6 outcomes
P( Sum 6) = 6 / 36 = 1 / 6
Sum 7 i.e., [(4,3), (4,3), (5,2), (5,2),(6,1), (6,1)] =6 outcomes
P( Sum 7) = 6 / 36 = 1 / 6
Sum 8 i.e., [(5,3), (5,3), (6,2), (6,2)] =4 outcomes
P( Sum 8) = 4 / 36 = 1 / 9
Sum 9 i.e., [(6,3), (6,3)] =2 outcomes.
P( Sum 9) = 2 / 36 = 1 / 18​​​

(i)There are 500 possible ways to draw a card.Perfect squares of even numbers are 4, 16, 36, 64, 64, 100, 144,196, 256, 324, 400, 484.
Number of ways to draw a no. of which is a perfect square of even number =11
Probability of first player winning a prize = \(11\over 500\)
(ii)For second players No. of cards left = 500-1=499
[∵ One card has been drawn by 1st players and it has not been replaced]
Also No. of cards bearing perfect square of even number = 11
[∵ 1st players has not won ∴ Card drawn by him does not bear a perfect square even numbert]
∴ Probability of IInd player winning a prize = \(11\over 499\)

Total outcomes
(H, H), (H, T), (T, 1), (T, 3), (T, 4), (T, 5), (T, 6)
A=(H, H), (H, T)
B=(T, 2), (T, 4), (T, 6)
C=(H, T), (T, 1), (T, 2), (T, 3), (T, 4), (T, 5), (T, 6)
D=(T, 1), (T, 3), (T, 5)

Possible outcomes on tossing a coin 3 times, are HHH, HHT, HTH, THH, HTT, THT, TTH, TTT
\(\therefore\)  Total number of outcomes = 8
(i) Let E₁ be the event that Sweta losses the entry fee
i.e. shen tosses tail three times i.e. TTT.
\(\therefore\) Number of outcomes favourable to E₁ = 1
Hence, required probability = P(E₁) = \(\frac{1}{8}\)
(ii) Let E₂, be the event that Sweta gets double entry fee
i.e. she tosses heads three times
i.e. HHH
\(\therefore\) Number of outcomes favourable to E₂ = 1
Hence, required probability \(=P\left(E_2\right)=\frac{1}{8}\)
(iii) Let E₃ be the event that Sweta gets her entry fee back
i.e. Sweta gets heads one or two times
i.e. event of getting
HTT, THT, TTH, HHT, HTH or THH
\(\therefore\) Number of outcomes favourable to E₃ = 6
Hence, required probability = P(E₃) = \(\frac{6}{8}=\frac{3}{4}\)

When these dice are thrown then total possible outcomes are
\(\therefore\)Total possible  outcomes = 36
(i)Sum of number = 7
Favourable outcomes are (3,4), (4, 3), (4, 3), (5, 2), (5,2), (6,1)
\(\therefore\)Favourable wavs = 6
\(\therefore\)Probability that sum of number is 7 = \(\frac{6}{36}=\frac{1}{6}\) 
(ii)Sum is a perfect square i.e., sum is 4 or 9 (\(\therefore\)Maximum sum is 10)
Favoutable outcomes are (1, 3), ( 1, 3), (2, 2), (2, 2), (3, 4), (6, 3), (6, 3) = 7 outcomes.
\(\therefore\)Probability = \(\frac {7}{36}\)