A twelve-sided die with sides labeled through will be rolled once. Each number is equally likely to be rolled. What is the probability of rolling a number greater than ? Write your answer as a fraction in simplest form.

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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]

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