Complete Question:
A twelve-sided die with sides labeled 1 through 12 will be rolled once. Each number is equally likely to be rolled, what is the probability of rolling a number greater than 10?
Answer:
[tex]P(T) = \frac{1}{6}[/tex]
Step-by-step explanation:
Given
Number of Sides = 12
Required
Probability of obtaining a side greater than 10
We start by listing out the sample space;
[tex]S = \{1,2,3,4,5,6,7,8,9,10,11,12\}[/tex]
[tex]n(S) = 12[/tex]
Next, we list out digits greater than 10; Represent this with T
[tex]T = \{11,12\}[/tex]
[tex]n(T) = 2[/tex]
Probability of T is calculated as follows;
[tex]P(T) = \frac{n(T)}{n(S)}[/tex]
[tex]P(T) = \frac{2}{12}[/tex]
Divide the numerator and denominator by 2
[tex]P(T) = \frac{1}{6}[/tex]
Hence, the required probability is [tex]\frac{1}{6}[/tex]