Answer: Option D. -> $\frac{1}{2}$
:
D

When you flip a coin, there are two outcomes either head or tail. So the probability of occurrence of the head is the same as the probability of occurrence of the tail that is $\frac{1}{2}$. 

Answer: Option B. -> 0.5
:
B

A coin was tossed 1500 times but 100 times result was not counted.
So, the total number of times result counted = 1400
Number of times head comes = 700
Number of times tail comes = 1400 - 700 = 700
Probability of occurrence of tail $=\frac{\text{Number of times tail comes}}{\text{Total number of times result counted}}$
Probability of occurrence of tail $=\frac{700}{1400}$
Probability of occurrence of tail = 0.5

Answer: Option C. -> 0.25
:
C and D

Total frequency of occurrence of 2, 3, 5 and 6 = 60.

Frequency of occurrence of 1 or 4 = 80 - 60 = 20.
Probability of occurrence of 1 or 4 = $\frac{\text{Frequency of occurrence of 1 or 4}}{\text{Total number of times die is thrown}}$

So, probability of occurrence of 1 or 4 = $\frac{20}{80}$ = $\frac{1}{4}$ = 0.25

Answer: Option B. -> 0 to 1 (inclusive)
:
B

.

Answer: Option A. -> $\frac{1}{3}$
:
A

Since the goldsmith can choose amongst A, B and C, he has 3 possibilities of which the favorable outcome is choosing C. Therefore, probability that he chooses C is $\frac{1}{3}$.

Answer: Option A. -> $\frac{1}{3}$
:

Out of the givne 5 bags, bag 2 and bag 4 have more than 18 chocolates.
Probability(E) $=\frac{\text{number of favorable events}}{\text{total number of events}}$
$\therefore$ Probablity of a bag containing more than 18 chocolates $=\frac{\text{number of bags containing more than 18 chocolates}}{\text{total number of bags}}=\frac{2}{5}=0.4$

Answer: Option B. -> $\frac{3}{11}$
:
B

Total number of alphabets in the word MISSIMMIPPI = 11
Number of times M comes = 3
Probability of picking M = $\frac{\text{Number of times M comes}}{\text{Total number of alphabets in the word MISSIMMIPPI}}$​
So, probability of its occurrence will be $\frac{3}{11}$.

Answer: Option B. -> 1
:
B

A coin is tossed 100 times and consider the case when you only get tail 100 times and no head. So, in all the trials you will have a tail.
Probability of getting a tail = $\frac{\text{Number of times tails appear}}{\text{Number of times a coin is tossed}}$
Probability of getting a tail​ = $\frac{100}{100}=1$

Answer: Option B. -> $\frac{1}{365}$
:
B

Number of days in a year = 365
Therefore, total possible outcomes = 365
If they have birthday on same day, then number of favourable outcomes =1
Therefore, required probability $=\frac{1}{365}$

Answer: Option A. -> $\frac{1}{36}$
:
A
Number of all possible outcomes = (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) and (6,6).
The number of all possible outcomes = 36.
There is only one case (5,5) when 5 comes up both the times.
$\text{Probability that 5 will come up both the times}=\frac{\text{Number of times 5 comes up both the times}}{\text{Number of all possible outcomes}}$
$\therefore$ P (5 will come up both the times) =$\frac{1}{36}$