A pair of fair, standard dice are rolled. What is the probability the sum of the dice is 5?

When two dice are rolled then total possible outcomes are : 6 $\times$ 6 = 36
Favorable cases : {(1, 4) ; (2, 3); (3, 2); (4, 1)}
No. of favorable outcomes = 4
$\mathrm{required} \mathrm{probability} = \frac{\mathrm{no}. \mathrm{of} \mathrm{favorable} \mathrm{outcomes}}{\mathrm{total} \mathrm{possible} \mathrm{outcomes}} = \frac{4}{36} = \frac{1}{9}$