Sample Space online exercise for - Free Printable

Problem: Sample Space Diagrams for Two Dice

We are tasked with analyzing the outcomes when a blue die and a red die are rolled, and their scores are added together. Let's solve each part step by step.

---

#### Step 1: Complete the Sample Space Diagram

The sample space diagram shows all possible outcomes when two dice are rolled. Each cell in the table represents the sum of the numbers on the blue die (rows) and the red die (columns).

| Blue Die \ Red Die | 1 | 2 | 3 | 4 | 5 | 6 |
|--------------------|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |

---

#### Step 2: Solve Each Part

##### (a) Complete the sample space diagram
The table above is the completed sample space diagram.

##### (b) What is the probability of scoring a 4?

To find the probability of scoring a 4, we count the number of outcomes that result in a sum of 4 and divide by the total number of outcomes.

- Outcomes that sum to 4: (1, 3), (2, 2), (3, 1)
- Total number of outcomes: \(6 \times 6 = 36\)

Thus, the probability is:
\[
P(4) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{3}{36} = \frac{1}{12}
\]

##### (c) What is the probability of scoring a 9?

To find the probability of scoring a 9, we count the number of outcomes that result in a sum of 9 and divide by the total number of outcomes.

- Outcomes that sum to 9: (3, 6), (4, 5), (5, 4), (6, 3)
- Total number of outcomes: \(36\)

Thus, the probability is:
\[
P(9) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{4}{36} = \frac{1}{9}
\]

##### (d) What is the probability of scoring 10 or more?

To find the probability of scoring 10 or more, we count the number of outcomes that result in a sum of 10, 11, or 12 and divide by the total number of outcomes.

- Outcomes that sum to 10: (4, 6), (5, 5), (6, 4)
- Outcomes that sum to 11: (5, 6), (6, 5)
- Outcomes that sum to 12: (6, 6)

Total favorable outcomes: \(3 + 2 + 1 = 6\)

Thus, the probability is:
\[
P(\geq 10) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}
\]

##### (e) What is the probability of scoring 5 or less?

To find the probability of scoring 5 or less, we count the number of outcomes that result in a sum of 2, 3, 4, or 5 and divide by the total number of outcomes.

- Outcomes that sum to 2: (1, 1)
- Outcomes that sum to 3: (1, 2), (2, 1)
- Outcomes that sum to 4: (1, 3), (2, 2), (3, 1)
- Outcomes that sum to 5: (1, 4), (2, 3), (3, 2), (4, 1)

Total favorable outcomes: \(1 + 2 + 3 + 4 = 10\)

Thus, the probability is:
\[
P(\leq 5) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{10}{36} = \frac{5}{18}
\]

##### (f) What is the probability of getting a double (same number on each die)?

To find the probability of getting a double, we count the number of outcomes where the numbers on both dice are the same and divide by the total number of outcomes.

- Doubles: (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)

Total favorable outcomes: \(6\)

Thus, the probability is:
\[
P(\text{Double}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{6}{36} = \frac{1}{6}
\]

##### (g) What is the probability of the total score being a prime number?

To find the probability of the total score being a prime number, we identify the prime numbers between 2 and 12 (the possible sums) and count the outcomes that result in those sums.

- Prime numbers: 2, 3, 5, 7, 11
- Outcomes that sum to 2: (1, 1)
- Outcomes that sum to 3: (1, 2), (2, 1)
- Outcomes that sum to 5: (1, 4), (2, 3), (3, 2), (4, 1)
- Outcomes that sum to 7: (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
- Outcomes that sum to 11: (5, 6), (6, 5)

Total favorable outcomes: \(1 + 2 + 4 + 6 + 2 = 15\)

Thus, the probability is:
\[
P(\text{Prime}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{15}{36} = \frac{5}{12}
\]

##### (h) What is the probability of the total score being a factor of 12?

To find the probability of the total score being a factor of 12, we identify the factors of 12 and count the outcomes that result in those sums.

- Factors of 12: 1, 2, 3, 4, 6, 12
- Outcomes that sum to 2: (1, 1)
- Outcomes that sum to 3: (1, 2), (2, 1)
- Outcomes that sum to 4: (1, 3), (2, 2), (3, 1)
- Outcomes that sum to 6: (1, 5), (2, 4), (3, 3), (4, 2), (5, 1)
- Outcomes that sum to 12: (6, 6)

Total favorable outcomes: \(1 + 2 + 3 + 5 + 1 = 12\)

Thus, the probability is:
\[
P(\text{Factor of 12}) = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{12}{36} = \frac{1}{3}
\]

---

Final Answers:

\[
\boxed{
\begin{aligned}
&\text{(b)} \ P(4) = \frac{1}{12} \\
&\text{(c)} \ P(9) = \frac{1}{9} \\
&\text{(d)} \ P(\geq 10) = \frac{1}{6} \\
&\text{(e)} \ P(\leq 5) = \frac{5}{18} \\
&\text{(f)} \ P(\text{Double}) = \frac{1}{6} \\
&\text{(g)} \ P(\text{Prime}) = \frac{5}{12} \\
&\text{(h)} \ P(\text{Factor of 12}) = \frac{1}{3}
\end{aligned}
}
\]