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

Explanation:

Step1: Recall Conditional Probability Formula

The formula for conditional probability is \( P(A|B) = \frac{P(A \cap B)}{P(B)} \), and also \( P(A \cap B) = P(B \cap A) \) (since intersection is commutative).

Step2: Analyze \( P(C \cap D) \)

We are given that 11 percent have both dogs and cats, so \( P(C \cap D) = 0.11 \). Also, since \( A \cap B = B \cap A \), \( P(C \cap D) = P(D \cap C) \), so these two statements are true.

Step3: Calculate \( P(D|C) \)

Using the conditional probability formula, \( P(D|C) = \frac{P(D \cap C)}{P(C)} \). We know \( P(D \cap C) = 0.11 \) and \( P(C) = 0.25 \). So \( P(D|C) = \frac{0.11}{0.25} = 0.44 \), so this statement is true.

Step4: Calculate \( P(C|D) \)

Using the conditional probability formula, \( P(C|D) = \frac{P(C \cap D)}{P(D)} \). We know \( P(C \cap D) = 0.11 \) and \( P(D) = 0.32 \). So \( P(C|D) = \frac{0.11}{0.32} \approx 0.34375
eq 0.78 \), so this statement is false. And since \( P(C|D) \approx 0.34375 \) and \( P(D|C) = 0.44 \), they are not equal, so that statement is false.

Answer:

• \( P(D | C) = 0.44 \)
• \( P(C \cap D) = 0.11 \)
• \( P(C \cap D) = P(D \cap C) \)