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}
\(P(E)=\frac { Number\ of\ favourable\ outcomes }{ Total\ number\ of\ outcomes } \)
\(P(E)=\frac { n(E) }{ n(S) } =\frac { 2 }{ 6 } =0.333...\)

A= {3, 6, 9, 12, 15, 18, 7, 14}
P(A) = \(\frac { 8 }{ 20 } =\frac { 2 }{ 5 } \).

A = {2, 3, 5, 7, 11, 13, 17, 19, 23}
P(A) = \(\frac{9}{25}\).

S = {Sunday Monday, Monday Tuesday, Tuesday Wednesday, Wednesday Thursday, Thursday Friday, Friday Saturday, Saturday Sunday}
n(S) = 7; n (A) = 2; P(A) =\(\frac{2}{7}\).

S = {HH, HT, TH, TT}
(i) probability of two heads = \(\frac { 1 }{ 4 } \)
(ii) probability of one head = \(\frac { 1 }{ 2 } \)
(iii) probability of at least one head = \(\frac { 3 }{ 4 } \)
(iv) probability of at most one head = \(\frac { 3 }{ 4 } \).

Probability of getting an even number = \(\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \)
Probability  of getting a multiple of 3 =\(\frac { 2 }{ 6 } \)
Probability of getting an even multiple of 3 = \(\frac { 1 }{ 6 } \)
Probability of getting an even number or
a multiple of 3 = \(\frac { 1 }{ 2 } +\frac { 2 }{ 6 } -\frac { 1 }{ 6 } \)
=\(\frac { 3+2-1 }{ 6 } =\frac { 4 }{ 6 } =\frac { 2 }{ 3 } \)
(ii) Probability of a number between 3 and 6 =\(\frac { 2 }{ 6 } =\frac { 1 }{ 3 } \).

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

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) = \(\frac { n(A) }{ n(S) } =\frac { 1 }{ 4 } \).

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) = \(\frac { n(A) }{ n(S) } =\frac { 3 }{ 6 } =\frac { 1 }{ 2 } \).

(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 = \(\frac { n(A) }{ n(S) } =\frac { 1 }{ 7 } \).