Question

dr. edwards is a veterinarian who sees only dogs and cats. in each appointment, he may or may not give the animal a vaccine. the two - way frequency table summarizes dr. edwardss 80 appointments last week.

	no vaccine	vaccine
dog	22	16

let vaccine be the event that a randomly chosen appointment (from the table) included a vaccine. let cat be the event that a randomly chosen appointment (from the table) involved a cat. find the following probabilities. write your answers as decimals. (if necessary, consult a list of formulas.)
(a) ( p(\text{vaccine}) = square )
(b) ( p(\text{cat and vaccine}) = square )
(c) ( p(\text{cat} mid \text{vaccine}) = square )

Explanation:

Step1: Find total appointments

First, we calculate the total number of appointments. The two - way table has the following values: for "No vaccine" and "Cat" it's 18, "No vaccine" and "Dog" it's 22, "Vaccine" and "Cat" it's 24, "Vaccine" and "Dog" it's 16. So the total number of appointments \(N=18 + 22+24 + 16=80\).

Step2: Calculate \(P(\text{vaccine})\)

The number of appointments with vaccine is the sum of vaccine - cat and vaccine - dog appointments. So \(n(\text{vaccine})=24 + 16 = 40\). The probability \(P(\text{vaccine})=\frac{n(\text{vaccine})}{N}=\frac{40}{80}=0.5\).

Step3: Calculate \(P(\text{cat and vaccine})\)

The number of appointments that are both cat and vaccine is 24. So \(P(\text{cat and vaccine})=\frac{n(\text{cat and vaccine})}{N}=\frac{24}{80} = 0.3\).

Step4: Calculate \(P(\text{cat}|\text{vaccine})\)

Using the formula for conditional probability \(P(A|B)=\frac{P(A\cap B)}{P(B)}\), here \(A = \text{cat}\) and \(B=\text{vaccine}\). We know that \(n(A\cap B) = 24\) and \(n(B)=40\). So \(P(\text{cat}|\text{vaccine})=\frac{n(\text{cat and vaccine})}{n(\text{vaccine})}=\frac{24}{40}=0.6\).

Answer:

(a) \(P(\text{vaccine}) = 0.5\)
(b) \(P(\text{cat and vaccine})=0.3\)
(c) \(P(\text{cat}|\text{vaccine}) = 0.6\)