(i) Pob. of less than 41
\(=\frac{8+12}{90}=\frac{20}{90}=\frac{2}{9}\)
(ii) Prob.of more than 50 \(=\frac{20+13+17+5}{90}=\frac{55}{90}=\frac{11}{18}\)
(iii)Prob. of marks between 41 and 80
\(=\frac{15+20+13+17}{90}=\frac{65}{90}=\frac{13}{18}\)

\(\frac { 193 }{ 400 } \)

\((i)\frac { 1 }{ 4 }\) 
\((ii)\frac { 8 }{ 15 }\) 
\((iii)\frac { 13 }{ 60 } \)

Total number of students = 200
(i) Number of students who like statistics = 135
Probability that a student chosen at random likes statistics=\(\frac { 135 }{ 200 } =\frac { 27 }{ 40 } \)
(ii) Number of students who do not like statistics=65
Probability that a student chosen at random does not like it= \(\frac { 65 }{ 200 } =\frac { 13 }{ 40 } \)
Aliter:
 Probability that a student chosen at random does not like statistics
= 1 - probability that a student chosen at random likes statistics
 =\(1-\frac { 27 }{ 40 } =\frac { 13 }{ 40 } \) 

Total number of students born in the year
= 3 + 4 + 2 + 2 + 5 + 1 + 2 + 6 + 3 + 4 + 4 + 4 = 40
Number of students born in August = 6
. Probability that a student of the class was born in August =\(\frac { 6 }{ 40 } =\frac { 3 }{ 20 } \)

Required probability=\(\frac { 265 }{ 500 } =0.53\)

Required probability=\(1-\frac { 79 }{ 200 } \)

Required probability=0

1,2,3,4,5,6

H,T

(a) Probability that a student selected is opts French language = \(\frac{12}{50}=\frac{6}{25}\)
(b) Probability that a student selected does not opt for Japanese = 1-selected student opts Japnese
(c) Probability that selected either opts for Sanskrit or for German = Prob. of student opts Sanskrti+ Prob. of students opts German
\(\therefore \)No of student who opted German 
= 5 - (14 + 08 + 12 + 6)
= 50-40
= 10
\(\therefore \)Prob. that selected student either opt for Sanskrit or for German
\(=\frac{14}{50}+\frac{10}{50}\)
\(=\frac{24}{50}=\frac{12}{25}\)

(a) The number of drivers in the age of 25-40 years and has more than 2 accidents in the year
= 15 + 10 = 25
(b) The number of drivers the age of whose is above 40 years and has accidents more than 1 but less than 3
= 13 + 17 = 30.

Total number of subjects = 5
(i) Number of subjects in which the student gets a first class = 4
Probability that the students gets a first class \(=\frac { 4 }{ 5 } \)
(ii) Number of subjects in which the student gets a distinction = 2
Probability that the student gets a distinction  \(=\frac { 2 }{ 5 } \)
(iii) Number of subjects in which the student gets marks between 70% and 80% = 1
Probability that the students gets marks between 70% and 80% \(=\frac { 1 }{ 5 } \) 

Number of two wheelers = 57
Number of three wheelers = 33
Number of four wheelers = 30
Total number of vehicles = 57 + 33 + 30 = 120
Number of vehicles that is not a four-wheeler = 57 + 33 = 90
Probability that the vehicle chosen at random is not a four-wheeler
 \(\frac { 90 }{ 120 } =\frac { 3 }{ 4 } \)

Total number of families = 1500
(i) at most one girl means 0 girl or 1 girl.
Number of families which have at most one girl
= Number of families which have 0 girl + Number of families which have 1 girl.
= 211 + 814 = 1015
Probability that it has at most one girl=\(\frac { 1015 }{ 1500 } =\frac { 203 }{ 300 } \)
(ii) at least one girl means 1 girls or 2 girls.
 Number of families which have at least one girl
= Number of families which have I girl + Number of families which have 2 girls.
= 814 + 475 = 1289
 Probability that it has at least one girl=\(\frac { 1289 }{ 1500 } \) 

\((i)\frac { 2 }{ 5 }\) 
\((ii)\frac { 3 }{ 5 } \) 
\((iii)\frac { 11 }{ 20 } \)

\((i)\frac { 21 }{ 100 }\) 
\((ii)\frac { 6 }{ 25 } \) 

The total number of unit tests held is 5.
The number of unit tests in which the student obtained more than 70% marks is 3.
So, P(scoring more than 70% marks) \(=\frac{3}{5}=0.6\)

Let us denote the events of getting two heads, one head and no head by E₁, E₂ and E₃, respectively. So,
\( P\left(E_{1}\right)=\frac{105}{500}=0.21 \)
\(P\left(E_{2}\right)=\frac{275}{500}=0.55 \)
\(P\left(E_{3}\right)=\frac{120}{500}=0.24 \)

\(\frac { 1 }{ 2 } \)