ap stats lesson 5.4 homework name
1. brett asked his classmates in his 1st hour whether they had siblings and whether they had a pet. he recorded the information in the two - way table shown, including his own information. suppose a student from bretts 1st hour is chosen at random.
define events ( x ): has a sibling and ( y ): has a pet.

| | has a pet | no pets | total |
|--|--|--|--|
| has a sibling | 13 | 11 | 24 |
| no siblings | 2 | 1 | 3 |
| total | 15 | 12 | 27 |

a. find ( p(x) ) and explain what this means in words.
b. find ( p(y^{c}) ) and explain what this means in words.
c. find ( p(x ext{ or } y) ) and explain what this means in words.
d. are the events ( x ) and ( y ) mutually exclusive? explain.

Question

ap stats lesson 5.4 homework name
1. brett asked his classmates in his 1st hour whether they had siblings and whether they had a pet. he recorded the information in the two - way table shown, including his own information. suppose a student from bretts 1st hour is chosen at random.
define events ( x ): has a sibling and ( y ): has a pet.

| | has a pet | no pets | total |
|--|--|--|--|
| has a sibling | 13 | 11 | 24 |
| no siblings | 2 | 1 | 3 |
| total | 15 | 12 | 27 |

a. find ( p(x) ) and explain what this means in words.
b. find ( p(y^{c}) ) and explain what this means in words.
c. find ( p(x 	ext{ or } y) ) and explain what this means in words.
d. are the events ( x ) and ( y ) mutually exclusive? explain.

Accepted Answer

### Part a
# Explanation:
## Step1: Identify total and favorable
Total students: 27. Students with sibling (X): 24.
## Step2: Calculate probability
$P(X) = \frac{\text{Number with sibling}}{\text{Total}} = \frac{24}{27} = \frac{8}{9} \approx 0.889$
## Explanation in words:
The probability a random student has a sibling is $\frac{8}{9}$, meaning ~88.9% of students have siblings.

# Answer: $P(X) = \boldsymbol{\frac{8}{9}}$ (or ~0.889); it means the chance a randomly chosen student has a sibling is $\frac{8}{9}$.

### Part b
# Explanation:
## Step1: Identify $Y^C$ (no pet) count
Total with no pet: 12 (from "No pets" total column).
## Step2: Calculate probability
$P(Y^C) = \frac{\text{Number with no pet}}{\text{Total}} = \frac{12}{27} = \frac{4}{9} \approx 0.444$
## Explanation in words:
The probability a random student has no pet is $\frac{4}{9}$, meaning ~44.4% of students have no pet.

# Answer: $P(Y^C) = \boldsymbol{\frac{4}{9}}$ (or ~0.444); it means the chance a randomly chosen student has no pet is $\frac{4}{9}$.

### Part c
# Explanation:
## Step1: Use addition rule
$P(X \text{ or } Y) = P(X) + P(Y) - P(X \text{ and } Y)$
Find $P(Y)$: $\frac{15}{27}$, $P(X \text{ and } Y)$: $\frac{13}{27}$ (students with sibling and pet).
## Step2: Substitute values
$P(X \text{ or } Y) = \frac{24}{27} + \frac{15}{27} - \frac{13}{27} = \frac{24 + 15 - 13}{27} = \frac{26}{27} \approx 0.963$
## Explanation in words:
The probability a random student has a sibling or a pet (or both) is $\frac{26}{27}$, meaning ~96.3% of students have a sibling, a pet, or both.

# Answer: $P(X \text{ or } Y) = \boldsymbol{\frac{26}{27}}$ (or ~0.963); it means the chance a randomly chosen student has a sibling, a pet, or both is $\frac{26}{27}$.

### Part d
# Brief Explanations:
Mutually exclusive events cannot occur simultaneously (i.e., $P(X \text{ and } Y) = 0$). From the table, 13 students have both a sibling (X) and a pet (Y), so $P(X \text{ and } Y) = \frac{13}{27}
eq 0$. Thus, X and Y are not mutually exclusive.
# Answer: No, events $ X $ and $ Y $ are not mutually exclusive. Because there are 13 students who have both a sibling (event $ X $) and a pet (event $ Y $), so $ P(X \text{ and } Y) = \frac{13}{27}
eq 0 $, meaning the events can occur at the same time.

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QUESTION IMAGE

Question

Explanation:

Part a

Step1: Identify total and favorable

Step2: Calculate probability

Step1: Identify $Y^C$ (no pet) count

Step2: Calculate probability

Answer:

Part b