Dice Probability: Product as Perfect Square

When two dice are thrown simultaneously, each die has 6 possible outcomes (numbers 1 through 6). The total number of possible outcomes is the product of the outcomes for each die.

Total possible outcomes = $6 \times 6 = 36$.

Identifying Favorable Outcomes

We need to find the outcomes where the product of the numbers on the two dice ($d_1 \times d_2$) is a perfect square. A perfect square is a number that is the square of an integer (e.g., 1, 4, 9, 16, 25, 36, ...).

Let's list the pairs $(d_1, d_2)$ such that $d_1 \times d_2$ is a perfect square, where $d_1, d_2 \in \{1, 2, 3, 4, 5, 6\}$:

• If the product is 1: (1, 1)
• If the product is 4: (1, 4), (4, 1), (2, 2)
• If the product is 9: (3, 3)
• If the product is 16: (4, 4)
• If the product is 25: (5, 5)
• If the product is 36: (6, 6)

The set of favorable outcomes is {(1, 1), (1, 4), (4, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)}. The total number of favorable outcomes is 8.

Calculating the Probability

Probability is calculated as the ratio of the number of favorable outcomes to the total number of possible outcomes.

$ P(\text{Product is a perfect square}) = \frac{\text{Number of Favorable Outcomes}}{\text{Total Number of Outcomes}} $

$ P(\text{Product is a perfect square}) = \frac{8}{36} $

Simplifying the fraction gives:

$ \frac{8}{36} = \frac{2 \times 4}{9 \times 4} = \frac{2}{9} $

The probability that the product of the numbers appearing on the top faces of the dice is a perfect square is 2/9.