7.
| | shorts | no shorts | total |
| hat | 10 | 50 | 60 |
| no hat | 30 | 10 | 40 |
| | 40 | 60 | 100 |

a. how many total outcomes are possible?

b. $p(\text{shorts}) = \frac{40}{100}$

c. $p(\text{hat}) = \frac{60}{100}$

d. $p(\text{shorts and hat}) = \frac{10}{100}$

e. $p(\text{shorts|hat}) = \frac{10}{60}$

f. are wearing shorts and wearing a hat independent events? why or why not?

Question

7.
|       | shorts | no shorts | total |
| hat    | 10     | 50        | 60    |
| no hat | 30     | 10        | 40    |
|        | 40     | 60        | 100   |

a. how many total outcomes are possible?

b. $p(\text{shorts}) = \frac{40}{100}$

c. $p(\text{hat}) = \frac{60}{100}$

d. $p(\text{shorts and hat}) = \frac{10}{100}$

e. $p(\text{shorts|hat}) = \frac{10}{60}$

f. are wearing shorts and wearing a hat independent events? why or why not?

Accepted Answer

### Part a
# Explanation:
## Step1: Identify total from table
The table's bottom - right cell (Total row and Total column) gives the total number of outcomes.
From the table, the total number of outcomes is 100.
# Answer:
100

### Part b
# Explanation:
## Step1: Recall probability formula
The probability of an event $A$, $P(A)=\frac{\text{Number of favorable outcomes for }A}{\text{Total number of outcomes}}$.
For $P(\text{Shorts})$, the number of favorable outcomes (people wearing shorts) is 40 (from the "Shorts" column total), and the total number of outcomes is 100.
So $P(\text{Shorts})=\frac{40}{100}=\frac{2}{5}$ (or 0.4).
# Answer:
$\frac{2}{5}$ (or 0.4 or $\frac{40}{100}$)

### Part c
# Explanation:
## Step1: Recall probability formula
The probability of an event $A$, $P(A)=\frac{\text{Number of favorable outcomes for }A}{\text{Total number of outcomes}}$.
For $P(\text{Hat})$, the number of favorable outcomes (people wearing a hat) is 60 (from the "Hat" row total), and the total number of outcomes is 100.
So $P(\text{Hat})=\frac{60}{100}=\frac{3}{5}$ (or 0.6).
# Answer:
$\frac{3}{5}$ (or 0.6 or $\frac{60}{100}$)

### Part d
# Explanation:
## Step1: Recall probability formula for intersection
The probability of the intersection of two events $A$ and $B$, $P(A\cap B)=\frac{\text{Number of outcomes in both }A\text{ and }B}{\text{Total number of outcomes}}$.
For $P(\text{Shorts and Hat})$, the number of outcomes where a person is wearing both shorts and a hat is 10 (from the cell where "Hat" row and "Shorts" column meet), and the total number of outcomes is 100.
So $P(\text{Shorts and Hat})=\frac{10}{100}=\frac{1}{10}$ (or 0.1).
# Answer:
$\frac{1}{10}$ (or 0.1 or $\frac{10}{100}$)

### Part e
# Explanation:
## Step1: Recall conditional probability formula
The conditional probability $P(A|B)=\frac{P(A\cap B)}{P(B)}$. For $P(\text{Shorts}|\text{Hat})$, we can also use the formula $P(\text{Shorts}|\text{Hat})=\frac{\text{Number of people with shorts and hat}}{\text{Number of people with hat}}$.
The number of people with shorts and hat is 10, and the number of people with hat is 60.
So $P(\text{Shorts}|\text{Hat})=\frac{10}{60}=\frac{1}{6}\approx0.167$.
# Answer:
$\frac{1}{6}$ (or approximately 0.167 or $\frac{10}{60}$)

### Part f
# Explanation:
## Step1: Recall independence condition
Two events $A$ and $B$ are independent if $P(A\cap B)=P(A)\times P(B)$.
We know $P(\text{Shorts})=\frac{40}{100} = 0.4$, $P(\text{Hat})=\frac{60}{100}=0.6$, and $P(\text{Shorts and Hat})=\frac{10}{100} = 0.1$.
Now, calculate $P(\text{Shorts})\times P(\text{Hat})=0.4\times0.6 = 0.24$.
But $P(\text{Shorts and Hat}) = 0.1$, and $0.1
eq0.24$.
So, wearing shorts and wearing a hat are not independent events because the product of their individual probabilities is not equal to the probability of their intersection.
# Answer:
No, they are not independent. Because $P(\text{Shorts})\times P(\text{Hat})
eq P(\text{Shorts and Hat})$ ( $0.4\times0.6 = 0.24
eq0.1$).

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QUESTION IMAGE

Question

Explanation:

Part a

Step1: Identify total from table

Step1: Recall probability formula

Step1: Recall probability formula

Answer:

Part b