Let's analyze the problem in the image step by step.

Game Description:

You are playing a game called "Roll and Win" where you:

1. Pick a number.
2. Roll two regular dice (each numbered 1 through 6).
3. Add the numbers that come up on the dice.
4. If the sum equals the number you picked, you win a point.

For example, if you choose 11 and roll a 6 and a 5, you win because 6 + 5 = 11.

Part (a): Sample Space

The sample space is the set of all possible outcomes when rolling two dice. Since each die has 6 faces, the total number of possible outcomes is: $6 \times 6 = 36$ The sample space is the set of all pairs (i, j), where $i$ is the outcome on die 1 and $j$ is the outcome on die 2. The sums of the two dice range from $2$ (when both dice show 1) to $12$ (when both dice show 6).

Part (b): Outcome Table

To analyze the outcomes, you can create a table with sums of the dice. For each combination of the two dice, the sum is calculated. Below is an example table of possible sums:

Notice that certain sums, like 7, appear more frequently than others, such as 2 or 12. Each pair (i, j) is equally likely, but not all sums are equally likely because there are more ways to make some sums than others.

Part (c): Probability of Specific Events

• $P(\text{even})$: Probability of rolling an even sum.
• $P(10)$: Probability of rolling a sum of 10.
• $P(15)$: Probability of rolling a sum of 15.

Even Sums:

The even sums from rolling two dice are: $2, 4, 6, 8, 10, 12$.

Now, let's count how many ways each sum can occur:

• 2 occurs 1 time: (1,1)
• 4 occurs 3 times: (1,3), (2,2), (3,1)
• 6 occurs 5 times: (1,5), (2,4), (3,3), (4,2), (5,1)
• 8 occurs 5 times: (2,6), (3,5), (4,4), (5,3), (6,2)
• 10 occurs 3 times: (4,6), (5,5), (6,4)
• 12 occurs 1 time: (6,6)

So, there are $1 + 3 + 5 + 5 + 3 + 1 = 18$ even sums. The probability of an even sum is: $P(\text{even}) = \frac{18}{36} = \frac{1}{2}$

Sum of 10:

As seen in the table, the sum of 10 can occur in 3 ways: (4,6), (5,5), and (6,4). So, the probability is: $P(10) = \frac{3}{36} = \frac{1}{12}$

Sum of 15:

There is no way to roll a sum of 15 with two dice, so: $P(15) = 0$

Part (d): Most Likely Sum

The most likely sum is 7, as it occurs the most frequently (6 times) in the table:

• (1,6)
• (2,5)
• (3,4)
• (4,3)
• (5,2)
• (6,1)

The probability of rolling a sum of 7 is: $P(7) = \frac{6}{36} = \frac{1}{6}$

Would you like more details on any of the parts or further explanation?

Here are 5 related questions for practice:

1. What is the probability of rolling a sum of 2 or 12?
2. What is the probability of rolling an odd sum?
3. How many different ways can you roll a sum of 8?
4. If you roll the dice twice, what is the probability of getting a sum of 6 both times?
5. How would the probabilities change if you used dice numbered from 1 to 8?

Tip: When working with probabilities in dice games, always count the total number of possible outcomes to build your probability fraction.