We have total number of colonies = 5
(i) Number of colonies in which more than 40 plants survived  = 3(B,C and E)
\(\therefore \) P(more than 40 plants survived in a colony)\(=\frac{3}{5}\)
(ii) Number of colonies in which less than 41 plants survived = 2(A and D)
P(less than 41 plants survived in colony)\(=\frac{2}{5}\)
(iii) In order to keep environment safe, we should grow more and more plants.

Let E be the event the batsman hits a boundary.
Then \(\overline E\) is the event the batsman does not hit a boundary.
P(E) = \(\frac{12}{48}=\frac{1}{4}\)
\(\Rightarrow P(\overline E)=1-P(E)\)
\(=1-\frac{1}{4}=\frac{3}{4}\)

Total no. of drivers = 2000
(i) Number of drivers who are 18-29 years old and have exactly 3 accidents in one year in 61
So, \(P=\frac{61}{2000}\)
(ii) Number of drivers having 30-50 years of age and having one or more accidents in one year.
= 125 + 60 + 22 + 18
= 225
So, probability, P\(=\frac{225}{2000}=\frac{9}{80}\)
(iii) Number of drivers having no accident in one year = 440 + 505 + 360 = 1305
So, probability P = \(\frac{1305}{2000}=\frac{261}{400}\)

Total student = 7 + 7 + 11 + 3 + 5 + 9 + 8 = 50
(i) P(height of student is more that 135 cm)\(=\frac{25}{50}=\frac{1}{2}\)
(ii)P(height of student is at least 145 cm)\(=\frac{17}{50}\)
(iii) p(height of student is less than 130 cm)\(=\frac{14}{50}=\frac{7}{25}\)
(iv) P(height of student is 125 or more but less than 140)\(=\frac{21}{50}\)

(i) Probability of less than 30 marks\(=\frac{7+10}{=90}=\frac{17}{90}\)
(ii) Probability of marks 60 or more marks\(=\frac{15+8}{90}=\frac{23}{90}\)
(iii) Probability of marks between 40 and 70\(=\frac{20+20+15}{90}\)
\(=\frac{55}{90}\)
(iv) Probability of marks 70 or more \(=\frac{8}{90}=\frac{4}{45}\)

Total number of cards in the deck when king, queen, and jack of clubs are removed
= 52 - 3 = 49
(a) Number of cards which are 'a heart' = 13
Probability of getting a heart = \(\frac { 13 }{ 49 } \)
(b) Number of cards which are 'a king' = 3
Probability of getting a king =\(\frac { 3 }{ 49 } \)
(c) Number of cards which are 'the 10 of hearts' = 1
Probability of getting 'the 10 of hearts' = \(\frac { 1 }{ 49 } \) 

(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 patients = 400
(i) Number of patients who were cured in less than 2 months = 210 + 105 = 315
 Probability that the patient will be cured in less than 2 months\(=\frac { 315 }{ 400 } =\frac { 63 }{ 80 } \)
(ii) Number of patients who were cured in 1 month or more but not more than 3 months
  = 105 + 60 = 165
Probability that the patient will be cured in 1 month or more but not more than 3 months \(=\frac { 165 }{ 400 } =\frac { 33 }{ 80 } \)

Total number of cards in the box = 100
(a) Numbers less than 14 are 2,3,4,5,6,7,8,9,10,11,12,13
Their number = 12
Probability that the number on the card is a number less than 14
\(=\frac { 12 }{ 100 } =\frac { 3 }{ 25 } \)
(b) Perfect square numbers are 4,9,16,25,36,49,64,81,100
Their number = 9
Probability that the number on the card is a number which is a perfect square
 \(=\frac { 9 }{ 100 } \)
(c) Prime numbers less than 20 are 2,3,5,7,11,13,17,19
Their number = 8
Probability that the number on the card is a prime number less than 20
\(=\frac { 8 }{ 100 } =\frac { 2 }{ 25 } \)

In the bag,
number of red balls = 5
number of white balls = 8
number of green balls = 4
number of black balls = 7
Total number of balls in the bag = 5 + 8 + 4 + 7 = 24
(i) Number of black balls = 7
Probability that the ball drawn is black =\(\frac { 7 }{ 24 } \) 
(ii) Number of balls that are not green = 5 + 8 + 7 = 20
Probability that the ball drawn is not green =\(\frac { 20 }{ 24 } \) =\(\frac { 5 }{ 6 } \)

Total number of days = 250
(i) Number of days on which the weather forecasts were correct = 175
Probability that on a given day it was correct = \(\frac { 175 }{ 250 } =\frac { 7 }{ 10 } \)
(ii) Probability that it was not correct on a given day=\(1-\frac { 7 }{ 10 } =\frac { 3 }{ 10 } \) 

P(likes the detergent) = \(\frac { 375 }{ 500 } \)
= \(\frac { 3 }{ 4 } \)
(ii) P(does not like the detergent)
= 1-p(likes the detergent)
= \(1-\frac { 3 }{ 4 } =\frac { 1 }{ 4 } \)