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