The sum of probabilities of two mutually exclusive events need not be always 1.
For example, consider the events “getting a prime number”, and “getting a composite number” for the experiment of rolling 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.) $=\frac{2}{6}=\frac{1}{3}$
$\frac{1}{2}+\frac{1}{3}\ne 1.$ So The sum of probabilities of two mutually exclusive events need not be always 1.

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

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

$P\left(\text{getting sum as 7}\right)=\frac{\text{Favourable outcomes}}{\text{Total possible outcomes}}$

                                $=\frac{Outcomehavingsamenumbers}{TotalOutcomes}$

                                $=\frac{6}{36}$ 

                                = $\frac{1}{6}$

The sum of values on rolling two dice can be any integer value between 2 and 12. Hence, there are 11 outcomes.

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

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.

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 $\frac{1}{4}$ .

Each country occupies 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

Given,

Sample space for rolling a pair of dice = S { (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) }

$\Rightarrow$ Total number of outcomes = 36

From the sample space, (1,6), (2,5), (3,4), (4,3), (5,2) and (6,1) gives sum of 7.

$\Rightarrow$  Number of favourable outcomes = 6

Probability of an event, P(E) = $\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$

P ( getting sum 7) =$\frac{6}{36}=\frac{1}{6}$
Therefore, the probability of getting a sum of 7 on rolling a pair of dice is $\frac{1}{6}$ .