Answer:
P(getting a 5)=[tex]\frac{1}{6}[/tex]
P(number larger than 2)=[tex]\frac{2}{3}[/tex]
P(numbers which are odd) =[tex]\frac{1}{2}[/tex]
Step-by-step explanation:
When a standard die is rolled, we get sample space as {1,2,3,4,5,6}
So, let's use formula
P(E)=[tex]\frac{outcomes favorable to event E}{Total number of outcomes}[/tex]
For part A:
Event is getting a 5.
So, out of 6 samples , we get one 5 as outcome
So,
P(getting a 5)=[tex]\frac{1}{6}[/tex]
For part B:
The numbers larger than 2 are {3,4,5,6}
So, p(number larger than 2)=[tex]\frac{4}{6} =\frac{2}{3}[/tex]
P(number larger than 2)=[tex]\frac{2}{3}[/tex]
For part C:
The numbers which are odd are {1,3,5}
So, p(numbers which are odd)=[tex]\frac{3}{6 } =\frac{1}{2}[/tex]
P(numbers which are odd) =[tex]\frac{1}{2}[/tex]