Question

a veterinarian surveys her clients and finds that 32 percent of the households have dogs, 25 percent have cats, and 11 percent have both dogs and cats. let event c be choosing a client who has cats and let event d be choosing a client who has dogs. which statements are true? check all that apply.

\\(p(c \mid d) = 0.78\\)
\\(p(d \mid c) = 0.44\\)
\\(p(c \cap d) = 0.11\\)
\\(p(c \cap d) = p(d \cap c)\\)
\\(p(c \mid d) = p(d \mid c)\\)

Explanation:

Identify given probabilities

Using the Probability Notation and Union and Intersection of Events knowledge points

$$
LATEXBLOCK0
$$

Evaluate intersection commutativity

Using the Union and Intersection of Events knowledge point

$$ P(C \cap D) = P(D \cap C) = 0.11 $$

Calculate conditional probabilities

Using the Conditional Probability knowledge point

$$
LATEXBLOCK1
$$

Compare conditional probabilities

Using the Conditional Probability knowledge point

$$ P(C \mid D) \approx 0.34 eq P(D \mid C) = 0.44 $$

Select true statements

Using the Probability Notation and Conditional Probability knowledge points

$$
LATEXBLOCK2
$$

Answer:

• \(P(C \mid D) = 0.78\)
• \(P(D \mid C) = 0.44\) (Correct answer)
• \(P(C \cap D) = 0.11\) (Correct answer)
• \(P(C \cap D) = P(D \cap C)\) (Correct answer)
• \(P(C \mid D) = P(D \mid C)\)