Total number of centum scorers = 1184
Therefore n = 1184
(i) Let E₁ be the event of getting a centum scorer in Mathematics.
Therefore n(E₁) = 233, That is, r₁ = 233
\(P({ E })_{ 1 }=\frac { { r }_{ 2 } }{ n } =\frac { 233 }{ 1184 } \)
(ii) Let E₂ be the event of getting a centum scorer in Science.
Therefore n(E₂ ) = 106, That is, r₂ = 106
\(P({ E })_{ 2 }=\frac { { r }_{ 2 } }{ n } =\frac {106 }{ 1184 } \)
P(E^'₂) = 1− P(E₂)
\(=1-\frac { 106 }{ 1184 } \)
\(=\frac { 1078 }{ 1184 } \)

Let E be the event that it will rain tomorrow. Then E′ is the event that it will not rain tomorrow.
Since P(E) = 0.91, we have P(E′) = 1−0.91 (how?)
= 0.09
Therefore, the probability that it will not rain tomorrow
= 0.09

In this experiment, each trial is a match where Team I faces Team II.
We are concerned about the winning status of Team I.
There are 10 trials in total; out of which Team I wins in the 1^st, 6^th and 9^th matches.
The relative frequency of Team I winning the matches = \(\frac { 3 }{ 10 } \)or 0.3

Total number of staff members = 42
The relative frequencies:
Car users \(=\frac { 7 }{ 42 } =\frac { 1 }{ 6 } \)
Two-wheeler users \(=\frac { 20 }{ 42 } =\frac { 10 }{ 21 } \)
Cycle users \(=\frac { 15 }{ 42 } =\frac { 5 }{ 14 } \)

Sample space S = {1, 2, 3, 4, 5, 6}
Let E be the event of getting a number greater than 4
E = {5, 6}
\(P(E)=\frac { Number\ of\ favourable\ outcomes }{ Total\ number\ of\ outcomes } \)
\(P(E)=\frac { n(E) }{ n(S) } =\frac { 2 }{ 6 } =0.333...\)

Total number of families S = 1500
n(S) = 1569
Let P be the event of selecting a family having part time maids and F be the event of selecting a family having full time maids.
(i) Both types of maids Let P \(\cap\) F be the event of selecting a family having both types of maids.
Let (P \(\cap\) F) = 259
\(\mathrm{P}(\mathrm{P} \cap \mathrm{F})=\frac{\mathrm{n}(\mathrm{P} \cap \mathrm{F})}{\mathrm{n}(\mathrm{S})}=\frac{250}{1500}=\frac{1}{6}\)
(ii) Part time maids Part time maids = only part time maid + both
= 860 + 250 = 1110 '
n(P) = 1110
\(\mathrm{p}(\mathrm{P})=\frac{\mathrm{n}(\mathrm{P})}{\mathrm{n}(\mathrm{S})}=\frac{1110}{1500}=\frac{111}{150}\)
(iii) No maids Let (P \(\cup\) P)'be the event of choosing a family not having maids and (P u F) be the event of choosing a family having part time or fuli time or both maids.
n(P \(\cup\) F ) = only n(P) + only n(F) + n(p \(\cap\) p) = 850 + 370 + 250
n(P\(\cup\)F) = 1480
n(P\(\cup\)F)' = n(S)-n(P\(\cup\)F)
= 1500 - 1480
n(P \(\cup\) F)' = 20
\(\mathrm{P}(\mathrm{P} \cup \mathrm{F})^{\prime}=\frac{\mathrm{n}(\mathrm{P} \cup \mathrm{F})^{\prime}}{\mathrm{n}(\mathrm{S})}=\frac{20}{1500}=\frac{1}{75}\)

p(A) = 0.72
P(A') = 1 - 0.72 = 0.28

\(\frac { x }{ 3 } +\frac { x }{ 5 } \) = 1
\(\frac { 5x+3x }{ 15 } \) = 1
\(\frac { 8x }{ 15 } \) = 1
8x = 15
x = \(\frac { 15 }{ 8 } \)

No. of youngsters = 400
n(S)
No. of youngsters having voter id = 191
n(A)
No. of youngsters do not have their voter id
n(B) = 400 - 191 = 209
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 209 }{ 400 } \).

Total n(S) = 10,000
Defective n(A) = 25
Number of defective laptop is 25
Number of good laptop is = 10000 - 25 = 9975
Let A be the event of choosing a good laptop. n(A) = 9975
\(\mathrm{P}(\mathrm{A})=\frac{\mathrm{n}(\mathrm{A})}{\mathrm{n}(\mathrm{S})}=\frac{9975}{10000}\)
P(A) = 0.9975
Probability of getting a good laptop is 0.9975

(i) What is the probability that the spinner will not land on a multiple of 2?
(ii) What is the probability that the spinner will land on an odd number?

Total no. of choices = 8
n(S)
Total no. of multiples of 3 (A) = {3, 6}
n(A) = 2
Event of non-multiples of 3(B) = {1, 2, 4,5, 7, 8}
n(B) = 6
\(\frac { n(B) }{ n(S) } =\frac { 6 }{ 8 } =\frac { 3 }{ 4 } \).

Total no. of attempts n(S) = 40
Total no. of attempts by A team n(A) = 32
Total no. of attempts by the opponent team B = n(B) = 40 - 32 = 8
P(B) = \(\frac { n(B) }{ n(S) } =\frac { 8 }{ 40 } =\frac { 1 }{ 5 } \).

n(S) = 7000 
S - Total no. of lights.
n(A) = 25   
A - Defective ones.
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 25 }{ 7000 } =\frac { 1 }{ 280 } \).

When two dice are rolled
Sample space
S = {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4),(4,5),(4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} 
n(S) = 36
(i) Event of the sum is equal to 1 = 0
∴ Probability = \(\frac { 0 }{ n(S) } \) = 0
(ii) Event of the sum is equal to 4
B = {(1, 3), (2, 2), (3, 1)}
n(B) = 3
P(B)= \(\frac { n(B) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 12 } \)
(iii) Event of the sum is equal to less than 13
C= {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)} 
n(C) = 36
P(C) = \(\frac { n(C) }{ n(S) } =\frac { 36 }{ 6 } \).

sample space when two coins are tossed (S) = {HH, TT, HT, TH}
n(S) = 4
Event of getting two heads (A) = {HH}
n(A) = 1
Probability of getting two heads p(A) = \(\frac { n(A) }{ n(S) } =\frac { 1 }{ 4 } \).

n(S) = 24
Red = n(R) = 3
Blue = n(B) = 5
Green= n(G) = 16
(i) Probability of picking a Blue ball =\(\frac { n(B) }{ n(S) } =\frac { 5 }{ 24 } \)
(ii) Probability of picking a Red ball =\(\frac { n(R) }{ n(S) } =\frac { 3 }{ 24 } =\frac { 1 }{ 8 } \)
(iii) Probability of picking a Green ball =\(\frac { n(G) }{ n(S) } =\frac { 16 }{ 24 } =\frac { 2 }{ 3 } \).

Faces of a dice (S) = {1, 2, 3, 4,5, 6}
n(S) = 6
Event of throwing an even number
A = {2, 4, 6}, n(A) = 3
∴ Probability of throwing an even number
P(A) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \).

Number of cards n(S) = 52
No. of King cards n(A) = 4
No. of Queen cards n(B) = 4
No. of Jack cards n(C) = 4
Probability of drawing a King card
\(\frac { n(A) }{ n(S) } =\frac { 4 }{ 52 } \)
Probability of drawing a Queen card
=\(\frac { n(B) }{ n(S) } =\frac { 4 }{ 52 } \)
Probability of drawing a Jack card
=\(\frac { n(C) }{ n(S) } =\frac { 4 }{ 52 } \)
∴ The Probability of drawing a King or a Queen or a Jack from a deck of cards
= p(A) + P(B)+ P(C) =\(\frac { 4 }{ 52 } +\frac { 4 }{ 52 } +\frac { 4 }{ 52 } =\frac { 4+4+4 }{ 52 } =\frac { 12 }{ 52 } =\frac { 3 }{ 13 } \).

(S) Days in a week = {Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday}
n(S) = 7
∴ No. of days in week = 7
(A) Event of selecting Sunday = {Sunday}
n(A) = 1
∴ Probability of selecting Sunday = \(\frac { n(A) }{ n(S) } =\frac { 1 }{ 7 } \).