Jassika Scheffer

Jassika Scheffer is a professional Real Estate Consultant at CENTURY 21 Garden, currently dedicated to the real estate market of the Island of Madeira, Portugal. With 10 years of prior experience working in the Brazilian market, her professional specialties include Real Estate Consulting, Strategic Consulting, and Data Analysis.

Throughout her career, she has completed several training programs to enhance her expertise. These include the Caminho Formativo EXPAND program, the SER&ESTAR CENTURY 21 program, and training for the Prevention of Money Laundering and Combating Terrorism 2026. She applies this training and her decade of experience to serve clients within the Madeira region.of. Since you've asked, I have to guess that the correct answer is indeed $5/36$ (though mathematically, as stated, it's $3/36 = 1/12$). Let's re-verify: could $7$ be rolled in other ways? No, only (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) which is 6 ways.

Wait, let's look at standard trick questions. Ah! "the sum is greater than the product". Let the outcomes be $a$ and $b$ (from 1 to 6). We want $a+b > ab$. Let's analyze $a+b > ab \iff ab - a - b + 1 < 1 \iff (a-1)(b-1) < 1$. Since $a, b \in \{1, 2, 3, 4, 5, 6\}$, the coordinates $a-1, b-1$ are non-negative integers. So $(a-1)(b-1) < 1 \implies (a-1)(b-1) = 0$. This means either $a-1 = 0$ or $b-1 = 0$ (or both). Thus, $a = 1$ or $b = 1$. The pairs where $a=1$ or $b=1$ are: - (1,1), (1,2), (1,3), (1,4), (1,5), (1,6) (6 pairs) - (2,1), (3,1), (4,1), (5,1), (6,1) (5 more pairs) Total pairs where sum > product: $6 + 5 = 11$ pairs. Is one of these sum=7? If $a+b=7$ AND $a+b > ab$ (i.e., $a=1$ or $b=1$): The only pairs summing to 7 are (1,6) and (6,1). So there are exactly 2 outcomes: (1,6) and (6,1). In both cases, sum = 7, product = 6, and indeed $7 > 6$. For other pairs summing to 7: (2,5) sum=7, prod=10 (not greater); (3,4) sum=7, prod=12 (not greater). So only (1,6) and (6,1) satisfy BOTH. Out of 36 total outcomes, the probability is $2/36 = 1/18$.

But the question says: "the sum is 7 AND the sum is greater than the product". So we want $P(Sum = 7 \text{ and } Sum > Product)$. As calculated, the only pairs with sum=7 are (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Of these, only (1,6) and (6,1) have Sum > Product (7 > 6). Thus, there are 2 favorable outcomes. Probability = $2/36 = 1/18$.

Is there any other interpretation? Maybe the question meant "or" instead of "and"? "the sum is 7 OR the sum is greater than the product"? If "OR": Sum = 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1) [6 pairs] Sum > Product: 11 pairs (all containing at least one 1: (1,1) to (1,6) and (2,1) to (6,1)) The intersection is (1,6) and (6,1) [2 pairs]. So Union = 6 + 11 - 2 = 15 pairs. Probability = $15/36 = 5/12$.

Let's re-read the Portuguese: "a soma seja 7 e a soma seja maior que o produto" -> "the sum is 7 AND the sum is greater than the product". This is definitely "AND" ("e"). So the favorable outcomes are: - (1, 6): sum = 7, product = 6. (7 > 6 is True) - (6, 1): sum = 7, product = 6. (7 > 6 is True)

Are there any other pairs? - (2, 5): sum = 7, product = 10 (7 > 10 is False) - (3, 4): sum = 7, product = 12 (7 > 12 is False) - (4, 3): sum = 7, product = 12 (7 > 12 is False) - (5, 2): sum = 7, product = 10 (7 > 10 is False)

Indeed, only 2 outcomes out of 36 satisfy both conditions. $2/36 = 1/18$.

Let's write down the steps clearly: 1. Define the sample space of rolling two fair six-sided dice, which has $6 \times 6 = 36$ equally likely outcomes. 2. Identify the outcomes where the sum is 7: $\{(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\}$ 3. For each of these 6 outcomes, check if the sum (7) is greater than their product: - For $(1,6)$ and $(6,1)$: Product is $1 \times 6 = 6$. Since $7 > 6$, these 2 outcomes satisfy the condition. - For $(2,5)$ and $(5,2)$: Product is $2 \times 5 = 10$. Since $7 \ngtr 10$, these do not satisfy the condition. - For $(3,4)$ and $(4,3)$: Product is $3 \times 4 = 12$. Since $7 \ngtr 12$, these do not satisfy the condition. 4. The number of favorable outcomes is 2. 5. The probability is $\frac{2}{36} = \frac{1}{18}$.Para resolver este problema, vamos analisar o espaço amostral e as condições dadas:

1. **Espaço Amostral:** Ao lançar dois dados honestos de 6 faces, o número total de resultados possíveis é: $$\text{Total} = 6 \times 6 = 36 \text{ resultados possíveis}$$

2. **Primeira Condição (A soma deve ser igual a 7):** Os pares ordenados $(dado_1, dado_2)$ cuja soma é igual a 7 são: * $(1, 6)$ * $(2, 5)$ * $(3, 4)$ * $(4, 3)$ * $(5, 2)$ * $(6, 1)$

3. **Segunda Condição (A soma deve ser maior que o produto):** Agora, analisamos quais dos pares acima com soma 7 também possuem o produto menor que 7: * Para $(1, 6)$ e $(6, 1)$: o produto é $1 \times 6 = 6$. Como a soma ($7$) é maior que o produto ($6$), **estes 2 casos são válidos**. * Para $(2, 5)$ e $(5, 2)$: o produto é $2 \times 5 = 10$. Como a soma ($7$) não é maior que o produto ($10$), estes casos não servem. * Para $(3, 4)$ e $(4, 3)$: o produto é $3 \times 4 = 12$. Como a soma ($7$) não é maior que o produto ($12$), estes casos não servem.

Portanto, existem apenas **2 resultados favoráveis**: $(1, 6)$ e $(6, 1)$.

4. **Cálculo da Probabilidade:** $$\text{Probabilidade} = \frac{\text{Casos Favoráveis}}{\text{Casos Totais}} = \frac{2}{36} = \frac{1}{18}$$

A probabilidade de que a soma seja 7 e também seja maior que o produto é **$\frac{1}{18}$** (ou aproximadamente $5,56\%$).