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 | d)=0.78 p(d | c)=0.44 p(c ∩ d)=0.11 p(c ∩ d)=p(d ∩ c) p(c | d)=p(d | c)

Explanation:

Step1: Recall the formula for conditional probability

The formula for conditional probability is \(P(A|B)=\frac{P(A\cap B)}{P(B)}\). Also, note that \(P(A\cap B) = P(B\cap A)\) (by the commutative property of intersection in probability).

Step2: Calculate \(P(C|D)\)

Given \(P(D) = 0.32\), \(P(C\cap D)=0.11\). Using the formula \(P(C|D)=\frac{P(C\cap D)}{P(D)}\), we substitute the values: \(P(C|D)=\frac{0.11}{0.32}\approx0.34\).

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

Given \(P(C) = 0.25\), \(P(C\cap D)=0.11\). Using the formula \(P(D|C)=\frac{P(C\cap D)}{P(C)}\), we substitute the values: \(P(D|C)=\frac{0.11}{0.25}=0.44\).

Step4: Analyze \(P(C\cap D)\)

We are given that 11 percent of households have both dogs and cats. So \(P(C\cap D) = 0.11\).

Step5: Analyze \(P(C\cap D)=P(D\cap C)\)

By the commutative property of intersection in probability, for any two events \(A\) and \(B\), \(P(A\cap B)=P(B\cap A)\). So \(P(C\cap D)=P(D\cap C)\) since \(C\cap D\) and \(D\cap C\) represent the same set of outcomes (households having both cats and dogs).

Step6: Analyze \(P(C|D)=P(D|C)\)

We found \(P(C|D)\approx0.34\) and \(P(D|C) = 0.44\). Since \(0.34
eq0.44\), \(P(C|D)
eq P(D|C)\).

Answer:

B. \(P(D|C) = 0.44\), C. \(P(C\cap D)=0.11\), D. \(P(C\cap D)=P(D\cap C)\)