A die is thrown once. Find the probability of getting

In a single throw of a die, the possible outcomes are 1, 2, 3, 4, 5 and 6.
Total number of possible outcomes = 6

(i)
Let E be the event of getting an even number.
Then, the favourable outcomes are 2, 4 and 6.
Number of favourable outcomes = 3

$\therefore$ Probability of getting an even number $=P(E)=\frac{\text { Number of favo u rable outcomes }}{\text { Total number of possible outcomes }}=\frac{3}{6}=\frac{1}{2}$

(ii)
Let E be the event of getting a number less than 5.
Then, the favourable outcomes are 1, 2, 3, 4
Number of favourable outcomes = 4

$\therefore$ Probability of getting a number less than $5=P(E)=\frac{\text { Number of favo u rable outcomes }}{\text { Total number of possible outcomes }}=\frac{4}{6}=\frac{2}{3}$

 (iii)
 Let E be the event of getting a number greater than 2.     
 Then, the favourable outcomes are 3, 4, 5 and 6.
 Number of favourable outcomes = 4

$\therefore$ Probability of getting a number greater than $2=P(E)=\frac{\text { Number of favo u rable outcomes }}{\text { Total number of possible outcomes }}=\frac{4}{6}=\frac{2}{3}$

(iv)
Let E be the event of getting a number between 3 and 6.

 Number of favourable outcomes = 2

$\therefore$ Probability of getting a number between 3 and $6=P(E)=\frac{\text { Number of favo u rable outcomes }}{\text { Total number of possible outcomes }}=\frac{2}{6}=\frac{1}{3}$

(v)
 Let E be the event of getting a number other than 3.  

 Number of favourable outcomes = 5

$\therefore$ Probability of getting a number other than $3=P(E)=\frac{\text { Number of favo u rable outcomes }}{\text { Total number of possible outcomes }}=\frac{5}{6}$

(vi)
Let E be the event of getting the number 5.  

 Number of favourable outcomes = 1

$\therefore$ Probability of getting the number $5=P(E)=\frac{\text { Number of favo u rable outcomes }}{\text { Total number of possible outcomes }}=\frac{1}{6}$