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

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

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

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

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) = \(\frac { n(A) }{ n(S) } =\frac { 25 }{ 7000 } =\frac { 1 }{ 280 } \).

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

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