Question

A die is thrown once, what is the probability of an odd number appears

Answer 1

Sample space = {1,2,3,4,5,6}
Total number of outcomes = 6
Favourable number of outcomes = 3
Therefore,
P(E)[where E is the event of getting an odd number]
= (Favourable outcomes)/(Total outcomes) = 3/6 = 1/2

Answer 2

Hi....
A die is thrown...
So sample space S = {1,2,3,4,5,6}
n(S) = 6
Let A be a event that a odd number appears
A={1,3,5}
n(A) = 3

So P(A) = n(A) /n(S) = 3/6 = 1/2

Hope this helps you.... ☺️☺️