(i) Number of days in a leap year=366 days
\(=(52\times 7+2)\) days=52 weeks and 2 days
Thus, a leap year always has 52 Fridays.
The remaining 2 days can be
(a) Sunday and Monday
(b) Monday and Tuesday
(c) Tuesday and Wednesday
(d) Wednesday and Thursday
(e) Thursday and Friday
(f) Friday and Saturday
(g) Saturday and Sunday.
Out of these 7 cases, we have Friday in two cases.
So, total number of possible outcomes, n(S)=7 and total number of favourable outcomes, n(E)=2
∴ P(53 Fridays)\(=\frac { n(E) }{ n(S) } =\frac { 2 }{ 7 } \)
(ii) Sample space of a die, S={1,2,3,4,5,6}
∴ n(S)=6
Let E=Event of getting an odd prime number
={3,5}
∴ n(E)=2
Hence, required probability\(=\frac { n(E) }{ n(S) } =\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \)

\((i){1\over26}(ii){}{1\over26}(iii){1\over4}(iv){3\over26}\)

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​​​

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) = \(\frac{6}{8}=\frac{3}{4}\)

\((i){10\over49}(ii){3\over49}\)

\((a) \frac{1}{2} \ \ \ \ (b) \frac{3}{43}\)

Total number of ways to draw ball from bag = 24
Number of ways to draw a white ball = x
\(\therefore \) Probability of drawing a white ball = \(\frac { x }{ 24 } \) 
A.T.Q., \(\frac { x }{ 24 } \)=y ....(i)
When 12 more white balls are added to the bag,
Number of balls in the bag = 24 + 12 = 36 Number of white balls in the bag = (x + 12)
\(\therefore \)Probability of drawing a white ball
=\(\frac { x+12 }{ 36 } \)
A.T.Q., \(\\ \frac { x+12 }{ 36 } =\frac { 5 }{ 3 } \times x\)
\(\Rightarrow \frac { x+12 }{ 36 } =\frac { 5 }{ 3 } \times \frac { x }{ 24 } \quad \{ using\quad (i)\} \)
\(\therefore \quad x+12=\frac { 36\times 5x }{ 72 } \)
\(\Rightarrow x+12=\frac { 35x }{ 2 } \)
\(\Rightarrow \) 2x+24=5x\(\Rightarrow \)3x=24\(\Rightarrow \)x=8

When a fair dice is rolled then possible outcomes are 1, 2, 3, 4, 5 or 6.

\(\therefore\)  Probability of x = 1.

P(sum is 8) = \(\frac{5}{36}\)

Total number of outcomes = 2³ = 8
(i) P(three heads) = \(\frac{1}{8}\)
(ii) P(atleast two tails) = \(\frac{4}{8}\)=\(\frac{1}{2}\)

When two coins are  tossed simultaneously, then possible outcomes are (H, H), (H, T), (T, H) and (T, T)
\(\therefore \) Total number of outcomes = 4
(i) Let E₁ be the event of getting exactly two head.
Then, the outcomes favourable to E₁ is (H, H).
 \(\therefore \) Number of outcomes favourable to E₁ = 1
\(\therefore \)  P(E₁) = \(\frac{1}{4}\)
(ii) Let E₂ be the event of getting atleast one head, i.e. one head or two head.
Then, favourable outcomes are (H, T), (H, H) and (H, T)
\(\therefore \)  Number of outcomes favourable to E₂ = 3
 \(\therefore \)  P (E₂ ) = \(\frac{3}{4}\)

\(\frac { 1 }{ 4 } \)

(i) Total angle = 360°
Angle of white = 120°
\(\therefore \text { Probability }=\frac{120^{\circ}}{360^{\circ}}=\frac{1}{3}\)
(ii) Total angle = 360°
Angle of blue = 60°
Angle of green = 60°
\(\therefore \text { Probability }=\frac{60^{\circ}+60^{\circ}}{360^{\circ}}=\frac{120^{\circ}}{360^{\circ}}=\frac{1}{3}\)
(iii) Let n students participated in the survey
\(\Rightarrow \frac{90^{\circ}}{360^{\circ}} \times n=15 \Rightarrow n=15 \times 4=60\)
\(\therefore\) 60 students participated in the survey.
Or
Total angle = 360°
Angle of red = 30°
Angle of blue = 60°
\(\therefore \text { Probability }=\frac{60^{\circ}+30^{\circ}}{360^{\circ}}=\frac{90^{\circ}}{360^{\circ}}=\frac{1}{4}\)

(i) (b):Total number of possible outcomes on throwing two dice simultaneously = 6 x 6 = 36
(ii) (c): As we know, that maximum sum of numbers on two dice = 6 + 6 = 12
It is an impossible event.
So, required probability = 0
(iii) (c) : Let A be the event that both the numbers on dice are prime.
 A = {(2, 2), (2, 3), (2, 5), (3,2), (3, 3), (3, 5), (5, 2), (5,3), (5, 5)}
\(\Rightarrow\)n(A) = 9
\(\begin{equation} P(A)=\frac{9}{36}=\frac{1}{4} \end{equation}\)
(iv) (c) : Let B the event that product of two numbers is odd.
B = {(1, 1), (1, 3), (1, 5), (3,1), (3, 3), (3, 5), (5,1), (5,3), (5, 5)}
\(\Rightarrow\)n(B) = 9
\(\begin{equation} P(B)=\frac{9}{36}=\frac{1}{4} \end{equation}\)
(v) (c) : Let C be the event that difference of two numbers is zero.
C = {(1, 1), (2,2), (3, 3), (4,4), (5, 5), (6, 6)}
\(\Rightarrow\)n(C) = 6
\(\begin{equation} P(C)=\frac{6}{36}=\frac{1}{6} \end{equation}\)