Question

A pair of fair dice is rolled. What is the probability that the second die lands on a higher value than does the first?

Short answer

$\frac{5}{12}$count the number of equally likely events.

Step-by-step solution

Given Information.

A pair of fair dice is rolled.

Explanation. 

Experiment: Two fair dice are rolled.

Question:$P(seconddielandsonabiggernumber)=?$

Since the dice are fair and distinct, each of the $6·6$rolls is equally likely.

To enumerate the number of possible configurations in the wanted event, choose two out of $6$the numbers - in $\left(\begin{array}{c}6\\ 2\end{array}\right)$ways. (Because each of these choices corresponds to one throw where the second dice gets a higher number)

$P\left(\text{second die scores higher}\right)=\frac{\left(\begin{array}{l}6\\ 2\end{array}\right)}{6·6}=\frac{15}{36}=\frac{5}{12}$

$\frac{5}{12}$count the number of equally likely events.