Total number of balls played = 30
Number of balls in which the batsman hit a boundary = 6 
Number of balls in which he did not hit a boundary
= 30 – 6
= 24
$\therefore \text{Probability of not hitting a boundary}=\frac{\text{Number of balls in which he does not hit a boundary}}{\text{Total number of balls played}}=\frac{24}{30}=0.8$

Number of parts exiting an assembly line which are bad(defective) = 8
Total number of parts passing the assembly line = 100
Probability of part being defective $=\frac{\text{Number of defective parts}}{\text{Total number of parts}}$
Probability of a part being defective $=\frac{8}{100}=0.08$

Total number of students in a class = 50
Number of students who passed an examination = 35
Probability that a student passed the exam = $\frac{\text{Number of students who passed an examinatio}}{\text{Total number of students in a class}}$
Probability that a student passed the exam = $\frac{35}{50}=0.7$.

The probability of an event always lies between 0 and 1 (0 and 1 inclusive).
Hence, among the options, only 0.001 can be the probability of an event.

Total number of outcomes when two dice are thrown =
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6) = 36 .
Sum of 12 = (6,6)
Therefore, Probability of getting a sum of 12 = $\frac{1}{36}$.

A six appears on a die 8 times out of 50 times.
Number of times six appears on the die = 8
Number of times the die is thrown = 50
Probability of getting a six $=\frac{\text{number of times six appears on the die}}{\text{total number of times the die is thrown}}$
Probability of getting a six $=\frac{8}{50}=0.16$

Number of good fruit = 120 – 24 = 96
Total number of fruits = 120
Probability of picking a good fruit $=\frac{\text{Number of good fruit}}{\text{Total number of fruits}}$
Probability of picking a good fruit $=\frac{96}{120}=0.8$.

Number of times  2 or 4 appears when a die was rolled = 175 + 150 = 325
Total number of times a die is thrown = 1000
Probability of getting a 2 or a 4 $=\frac{\text{number of times 2 or 4 appears}}{\text{total number of times a die is thrown}}$
$\Rightarrow \text{Probability}=\frac{325}{1000}=0.325$

In probability theory, the sample space of an experiment or random trial is the set of all possible outcomes or results of that experiment. 
For example, if we toss a single coin then the chances are that either head will appear or tail will appear, the sample space will be {H}, {T}.

There are always two possibilities with an event, either it will happen or it will not happen.
We know that the sum of probabilities of all possible outcomes of an event is 1.
$\therefore$ The sum of probabilities of an event happening and an event not happening = 1
Let P(E) be the probability of an event happening.
(Then, P(not E) is the probability of the event not happening)
Then, P(E) = 0.13  (Given)
$\Rightarrow$ P(not E) = 1 – 0.13 = 0.87