Two different dice are rolled together. The probability that the product of the numbers appeared is less than 18 is:

1. $\Big(\dfrac{1}{2}\Big)$

2. $\Big(\dfrac{2}{3}\Big)$

3. $\Big(\dfrac{7}{9}\Big)$

4. $\Big(\dfrac{13}{18}\Big)$

When two dice are thrown simultaneously, each die has 6 possible outcomes.

Total number of outcomes = 36

The pairs with a product ≥ 18 = {(3, 6), (4, 5), (4, 6), (5, 4), (5, 5), (5, 6), (6, 3), (6, 4), (6, 5), (6, 6)}

Total number of outcomes where product ≥ 18 = 10

∴ Total number of outcomes where product is less than 18 = 36 - 10 = 26.

Let E be the event that the product is less than 18, then

The number of favorable outcomes to the event E = 26

$\therefore P(E) = \dfrac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \dfrac{26}{36} = \dfrac{13}{18}$

Hence, option 4 is the correct option.