Answer: Option D. -> 8
:
D
$\text{Given: P(green marble)}=\frac{2}{3}$
Since there are 24 marbles, therefore the sample space will be 24 (i.e. the denominator must be 24.)
$\text{Let us assume there are}\text{x}\text{numberof}\text{green marbles in the jar.}$
$\text{For an event E, probability}P\left(E\right)=\frac{\text{Number of outcomes favorable to E}}{\text{Number of all possible outcomes of the experiment}}$
$\Rightarrow \text{P(green marbles)}=\frac{2}{3}=\frac{\text{x}}{24}$
$\Rightarrow \text{x}=16$
It implies that there are 16 green marbles in a jar of 24 marbles.
$\therefore \text{Numberofblue marbles}=24-16=8.$

Answer: Option D. -> 1
:
D
Total number of trials = 5
Number of trials which resulted in'6' = 5
$\therefore \text{Experimental Probability, P(E)}$
$=\frac{\text{Number of trials in which the event happened}}{\text{Total number of trials}}$
$=\frac{5}{5}$
$=1$

Answer: Option A. -> 0
:
A
The highest sum possible when rolling a pair of dice is 12(getting 6 on both dies). So getting a sum of 13 is an impossible event. Hence, the probability is 0.

Answer: Option D. -> 0.50
:
D
Each countryoccupies an area of 50$k{m}^2$
Probability that the person lands on a land occupied by Indians or Europeans = $\frac{(50+50)}{200}$ = $\frac{1}{2}$ = 0.50

:
The word TREKKING has 8 characters of which 2 are favourable. Hence, the probability is $\frac{2}{8}$= $\frac{1}{4}$.

Answer: Option B. -> False
:
B
The sum of probabilities of two mutually exclusive events need not be always1.
For example, consider the events “getting a prime number”, and “getting a composite number” for the experiment ofrolling a die. They are mutually exclusive, but the sum of their probabilities is less than 1.
Favorable outcomes for prime no. = 2 , 3, 5
Favorable outcomes for composite no. = 4, 6
P(getting a prime no.) $=\frac{3}{6}=\frac{1}{2}$
P(getting a composite no.) $=\frac{2}{6}=\frac{1}{3}$
$\frac{1}{2}+\frac{1}{3}\ne 1.$ SoThe sum of probabilities of two mutually exclusive events need not be always1.

Answer: Option B. -> 45
:
B
Given, total number of glasses $=25(20+5)$
⇒ Total number of outcomes $=25$
Let $E$ be the event of getting good glass.
Number ofgood ones $=20$
⇒ Number of favourable outcomes $=20$
$\text{Probability P(E)}=\frac{\text{Number of outcomes favourable to E}}{\text{Number of all possible outcomes of the experiment}}$
$=\frac{20}{25}=\frac{4}{5}$
$\therefore$ The probability of getting a good glass is $\frac{4}{5}$.

Answer: Option C. -> 16
:
C
The total possible outcomes when a die is rolled twice are:
$$\begin{array}{ccccccc}& \text{1}& \text{2}& \text{3}& \text{4}& \text{5}& \text{6}\\ 1& 1,1& 1,2& 1,3& 1,4& 1,5& 1,6\\ 2& 2,1& 2,2& 2,3& 2,4& 2,5& 2,6\\ 3& 3,1& 3,2& 3,3& 3,4& 3,5& 3,6\\ 4& 4,1& 4,2& 4,3& 4,4& 4,5& 4,6\\ 5& 5,1& 5,2& 5,3& 5,4& 5,5& 5,6\\ 6& 6,1& 6,2& 6,3& 6,4& 6,5& 6,6\end{array}$$
Total number of outcomes = 36
Number of ways to get sum as 7= {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
= 6
$P\left(\text{getting sum as 7}\right)=\frac{\text{Favourable outcomes}}{\text{Total possible outcomes}}$
$=\frac{Outcomehavingsamenumbers}{TotalOutcomes}$
$=\frac{6}{36}$
=$\frac{1}{6}$

Answer: Option C. -> 14
:
C
Given,
Total number of outcomes = 52
Number of diamonds = 13
Probability of an event, P(E) = $=\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$
P (getting a card of diamond) = $\frac{13}{52}=\frac{1}{4}$
Therefore, probability of getting a card of diamond is $\frac{1}{4}$ .

Answer: Option B. -> sample space
:
B
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 willappear or tail will appear,the sample space will be {H}, {T}.