To solve the problem, we need to analyze each scenario step by step. Let's go through them.

1. The probability that the sum of the numbers is less than or equal to 7

When two dice are thrown, the possible outcomes can be represented as pairs $(x, y)$, where $x$ and $y$ are the numbers shown on the first and second dice, respectively. Since each die has 6 sides, there are a total of $6 \times 6 = 36$ possible outcomes.

Next, let's find the combinations where the sum $x + y \leq 7$.

Combinations where $x + y \leq 7$:

• When $x = 1$, possible values of $y$ are: 1, 2, 3, 4, 5, 6 (6 outcomes)
• When $x = 2$, possible values of $y$ are: 1, 2, 3, 4, 5 (5 outcomes)
• When $x = 3$, possible values of $y$ are: 1, 2, 3, 4 (4 outcomes)
• When $x = 4$, possible values of $y$ are: 1, 2, 3 (3 outcomes)
• When $x = 5$, possible values of $y$ are: 1, 2 (2 outcomes)
• When $x = 6$, possible values of $y$ are: 1 (1 outcome)

Thus, the total number of outcomes where the sum is less than or equal to 7 is:

$6 + 5 + 4 + 3 + 2 + 1 = 21$

The probability is then the ratio of favorable outcomes to the total outcomes:

$P(\text{sum} \leq 7) = \frac{21}{36} = \frac{7}{12}$

2. The probability that at least one die shows a 3

To find the probability that at least one die shows a 3, we first calculate the complementary event, which is the probability that neither die shows a 3. If no die shows a 3, each die can show any of the numbers 1, 2, 4, 5, 6. So, each die has 5 possible outcomes, and the total number of outcomes where no die shows a 3 is:

$5 \times 5 = 25$

Thus, the number of favorable outcomes where at least one die shows a 3 is:

$36 - 25 = 11$

The probability is:

$P(\text{at least one die shows a 3}) = \frac{11}{36}$

3. The probability that at least one die shows a 3, given that the sum is less than 8

This is a conditional probability problem. We need to find $P(\text{at least one die shows a 3} \mid \text{sum} < 8)$, which is calculated using:

$P(A \mid B) = \frac{P(A \cap B)}{P(B)}$

Where:

• $A$ is the event "at least one die shows a 3"
• $B$ is the event "sum is less than 8"

We already know $P(B)$, the probability that the sum is less than 8, from part 1:

$P(B) = \frac{21}{36} = \frac{7}{12}$

Next, we find $P(A \cap B)$, which is the probability that the sum is less than 8 and at least one die shows a 3. From the list of sums less than or equal to 7, the combinations where at least one die is a 3 are:

$(1, 3), (2, 3), (3, 1), (3, 2), (3, 3), (3, 4), (4, 3), (5, 3), (3, 5)$

There are 9 such outcomes. Therefore, the probability of both events happening is:

$P(A \cap B) = \frac{9}{36} = \frac{1}{4}$

Now we can calculate the conditional probability:

$P(A \mid B) = \frac{\frac{1}{4}}{\frac{7}{12}} = \frac{1}{4} \times \frac{12}{7} = \frac{3}{7}$

Final Answers:

1. The probability that the sum is less than or equal to 7 is $\frac{7}{12}$.
2. The probability that at least one die shows a 3 is $\frac{11}{36}$.
3. The probability that at least one die shows a 3, given that the sum is less than 8, is $\frac{3}{7}$.

Would you like more details or have any questions?

Here are 5 related questions to explore:

1. What is the probability that the sum of the two dice is exactly 7?
2. What is the probability that both dice show the same number?
3. What is the probability that the sum is greater than 9?
4. How does the probability change if the dice are not fair?
5. What is the probability of getting at least one 6?

Tip: For conditional probability, always identify the total possible outcomes within the given condition before finding the favorable ones.