Respuesta :

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]