Probability Models

1. Since all four sections are equal, this is a uniform probability model — each outcome has the same probability.

$P(\text{red}) = P(\text{blue}) = P(\text{green}) = P(\text{yellow}) = \frac{1}{4}$

Verification:

$\frac{1}{4} + \frac{1}{4} + \frac{1}{4} + \frac{1}{4} = \frac{4}{4} = 1 \checkmark$

2. $P(\text{red}) = \frac{1}{2}, \quad P(\text{blue}) = \frac{1}{3}, \quad P(\text{green}) = \frac{1}{6}$

This is a non-uniform probability model because the outcomes do not have equal probabilities.

Verification:

$\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3}{6} + \frac{2}{6} + \frac{1}{6} = \frac{6}{6} = 1 \checkmark$

3. The sample space has $2 \times 6 = 12$ outcomes:

| | 1 | 2 | 3 | 4 | 5 | 6 | |---|---|---|---|---|---|---| | H | H1 | H2 | H3 | H4 | H5 | H6 | | T | T1 | T2 | T3 | T4 | T5 | T6 |

Outcomes with heads AND a number greater than $4$: H5, H6.

$P(\text{heads and} > 4) = \frac{2}{12} = \frac{1}{6}$

4. Tree diagram structure:

• Soup → Chicken, Fish, Beef, Pasta (4 meals)
• Salad → Chicken, Fish, Beef, Pasta (4 meals)
• Bread → Chicken, Fish, Beef, Pasta (4 meals)

Total number of meals: $3 \times 4 = 12$.

Meals with soup: Soup-Chicken, Soup-Fish, Soup-Beef, Soup-Pasta — that is $4$ meals.

$P(\text{soup}) = \frac{4}{12} = \frac{1}{3}$

5. a) Each die has $6$ faces, so the total number of outcomes is:

$6 \times 6 = 36$

b) Outcomes where the sum is $7$: $(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)$ — that is $6$ outcomes.

$P(\text{sum} = 7) = \frac{6}{36} = \frac{1}{6}$

c) Outcomes where the sum is greater than $9$ (i.e., sum $= 10, 11,$ or $12$):

• Sum $= 10$: $(4,6), (5,5), (6,4)$ — $3$ outcomes
• Sum $= 11$: $(5,6), (6,5)$ — $2$ outcomes
• Sum $= 12$: $(6,6)$ — $1$ outcome

Total: $3 + 2 + 1 = 6$ outcomes.

$P(\text{sum} > 9) = \frac{6}{36} = \frac{1}{6}$