1. in a community 70% of households have kids. of the households with kids, 60% have a pet. and of the households that dont have kids, 50% have a pet. make a tree diagram of this situation.
(a) p(household has kids and has a pet)
(b) p(household that does not have kids and does have a pet)
(c) p(household has a pet)
2. one hospital employs 200 nurses. nurses at the hospital can work either in the neonatal unit or as a pediatrician. 75% of all the nurses are female. 55% of all the female nurses work in the neonatal unit. 5% of male nurses work as a pediatrician. make a tree diagram of this situation.
(a) p(male)
(b) p(female and neonatal)
(c) p(male and pediatrician)
(d) p(pediatrician)

Question

Accepted Answer

### Problem 1 (Household and Pet Probability)

#### Part (a)
# Explanation:
## Step1: Identify Probabilities
Let $ K $ be the event that a household has kids, $ P $ be the event that a household has a pet. We know $ P(K) = 0.7 $ and $ P(P|K) = 0.6 $.

## Step2: Apply Multiplication Rule
The multiplication rule for conditional probability is $ P(K \cap P) = P(K) \times P(P|K) $. Substitute the values: $ 0.7 \times 0.6 = 0.42 $.

# Answer:
$ 0.42 $

#### Part (b)
# Explanation:
## Step1: Identify Probabilities
Let $ K^c $ be the event that a household does not have kids. $ P(K^c) = 1 - 0.7 = 0.3 $, and $ P(P|K^c) = 0.5 $.

## Step2: Apply Multiplication Rule
Using $ P(K^c \cap P) = P(K^c) \times P(P|K^c) $, substitute: $ 0.3 \times 0.5 = 0.15 $.

# Answer:
$ 0.15 $

#### Part (c)
# Explanation:
## Step1: Use Total Probability
The event $ P $ can occur in two ways: with kids ($ K \cap P $) or without kids ($ K^c \cap P $). We found $ P(K \cap P) = 0.42 $ and $ P(K^c \cap P) = 0.15 $.

## Step2: Sum the Probabilities
$ P(P) = P(K \cap P) + P(K^c \cap P) = 0.42 + 0.15 = 0.57 $.

# Answer:
$ 0.57 $

### Problem 2 (Nurse Probability)

#### Part (a)
# Explanation:
## Step1: Identify Probabilities
Let $ F $ be female, $ M $ be male. Total nurses $ N = 200 $. $ P(F) = 0.75 $, so $ P(M) = 1 - 0.75 = 0.25 $.

## Step2: Calculate $ P(M) $
$ P(M) = 0.25 $ (or $ \frac{50}{200} $ since $ 200 \times 0.25 = 50 $).

# Answer:
$ 0.25 $ (or $ \frac{1}{4} $)

#### Part (b)
# Explanation:
## Step1: Identify Probabilities
$ P(F) = 0.75 $, $ P(\text{Neonatal}|F) = 0.55 $.

## Step2: Apply Multiplication Rule
$ P(F \cap \text{Neonatal}) = P(F) \times P(\text{Neonatal}|F) = 0.75 \times 0.55 = 0.4125 $.

# Answer:
$ 0.4125 $

#### Part (c)
# Explanation:
## Step1: Identify Probabilities for Male
$ P(M) = 0.25 $, $ P(\text{Pediatrician}|M) = 0.05 $.

## Step2: Apply Multiplication Rule
$ P(M \cap \text{Pediatrician}) = P(M) \times P(\text{Pediatrician}|M) = 0.25 \times 0.05 = 0.0125 $.

# Answer:
$ 0.0125 $

#### Part (d)
# Explanation:
## Step1: Find Probabilities for Pediatrician
- For females: $ P(F \cap \text{Pediatrician}) = P(F) \times (1 - P(\text{Neonatal}|F)) = 0.75 \times (1 - 0.55) = 0.75 \times 0.45 = 0.3375 $.
- For males: $ P(M \cap \text{Pediatrician}) = 0.0125 $ (from part c).

## Step2: Sum the Probabilities
$ P(\text{Pediatrician}) = 0.3375 + 0.0125 = 0.35 $.

# Answer:
$ 0.35 $

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QUESTION IMAGE

Question

Explanation:

Problem 1 (Household and Pet Probability)

Part (a)

Step1: Identify Probabilities

Step2: Apply Multiplication Rule

Step1: Identify Probabilities

Step2: Apply Multiplication Rule

Step1: Use Total Probability

Step2: Sum the Probabilities

Answer:

Part (b)