Practice 2 Solutions

A’ ⋃ B

A ⋃ B’

3. A’

4. A ∩ B’

5. $\mathrm{P}(\mathrm{has} \mathrm{diploma})$

$41\%+18\%$

59%

6. The number of people who have a diploma and no experience

$$\begin{gathered} 18\% \mathrm{of} 150 \\ (0.18)\left(150\right)=27 \end{gathered}$$

There are 27 people with a diploma and no work experience.

7. The number of people who have a diploma or experience

$$\begin{gathered} 41\%+18\%+23\%=82\% \\ 82\% \mathrm{of} 150 \\ (0.82)\left(150\right)=123 \end{gathered}$$

There are 123 people who have a diploma or work experience.

8. $\mathrm{P}(\mathrm{no} \mathrm{diploma} \mathrm{and} \mathrm{no} \mathrm{experience})$

$\mathrm{P}(\mathrm{no} \mathrm{diploma} \cap \mathrm{no} \mathrm{experience})$

18%

9. Create a two-way table for rational, ℚ, and irrational, 𝕀, numbers.

Rational, ℚ: $–7, –\frac{1}{3}, 3,\frac{ 11}{2}, 12$

Irrational, 𝕀:$\sqrt{5}, e, \mathrm{\pi }$

Not rational or irrational: $–20i,\frac{ 5i}{6}, 13i$

Rational and irrational: none

10. Create a Venn diagram for rational, ℚ, and irrational, 𝕀, numbers.

Rational, ℚ: $–7, –\frac{1}{3}, 3,\frac{ 11}{2}, 12$

Irrational, 𝕀:$\sqrt{5}, e, \mathrm{\pi }$

Not rational or irrational: $–20i,\frac{ 5i}{6}, 13i$

Rational and irrational: none

11. $\mathrm{P}(\mathrm{\mathbb{Q}}’)$

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{\mathbb{Q}}’\right)& =& 1–\mathrm{P}\left(\mathrm{\mathbb{Q}}\right)\\ & =& 1–\frac{5}{11}\end{array}$

$\frac{6}{11}$

12. $\mathrm{P}\left(\mathrm{𝕀}\right)’$

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{𝕀}\right)’& =& 1–\mathrm{P}\left(\mathrm{𝕀}\right)\\ & =& 1–\frac{3}{11}\end{array}$

$\frac{8}{11}$

13. A rational number

$\mathrm{P}\left(\mathrm{\mathbb{Q}}\right)=\frac{5}{11}$

$\frac{5}{11}$

14. A rational and irrational number

$\mathrm{P}\left(\mathrm{\mathbb{Q}} \cap \mathrm{𝕀}\right)=0$

0

15. A sum of 5

$\mathrm{P}\left(\mathrm{sum} \mathrm{of} 5\right)=\frac{ 4}{36}$

$\frac{1}{9}$

16. A sum of 3 or 11

$\mathrm{P}\left(\mathrm{sum} \mathrm{of} 3 \cup \mathrm{sum} \mathrm{of} 11\right)=\frac{2}{36}+\frac{2}{36}=\frac{4}{36}$

$\frac{1}{9}$

17. $\mathrm{P}(\mathrm{sum} \mathrm{of} 9)’$

$\begin{array}{rcl}1–\mathrm{P}\left(\mathrm{sum} \mathrm{of} 9\right)& =& 1–\frac{4}{36}\\ & =& 1–\frac{1}{9}\end{array}$

$\frac{8}{9}$

18. $\mathrm{P}(\mathrm{sum} \mathrm{of} 6)$

$\frac{5}{36}$

19. Not the sum of either 12 or 3

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{sum} \mathrm{of} 12 \cup \mathrm{sum} \mathrm{of} 3\right)’& =& 1–\mathrm{P}\left(\mathrm{sum} \mathrm{of} 12 \cup \mathrm{sum} \mathrm{of} 3\right)\\ & =& 1–\left(\frac{1}{36}+\frac{2}{36}\right)\\ & =& 1–\frac{3}{36}=\frac{33}{36}\end{array}$

$\frac{11}{12}$

20. What sum is most likely to be rolled? Explain.

$\begin{array}{rcl}\mathrm{P}\left(\mathrm{sum} \mathrm{of} 7\right)& =& \frac{6}{36}=\frac{1}{6}\end{array}$

A sum of 7 occurs the most times, which means it has the greatest likelihood of being rolled.