4
AssignmentMultiple Select

Precision in Measurement

Question 4 • AZ-Common Core Algebra II A

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) = 0.78P(D | C) = 0.44P(C ∩ D) = 0.11P(C ∩ D) = P(D ∩ C)P(C | D) = P(D | C)

Answer
A
P(C | D) = 0.78
B
P(D | C) = 0.44
C
P(C ∩ D) = 0.11
D
P(C ∩ D) = P(D ∩ C)
E
P(C | D) = P(D | C)

Explanation

Use the conditional probability rule: P(C | D) = P(C ∩ D) / P(D) = 0.11 / 0.32 ≈ 0.34, so it is not 0.78. Likewise, P(D | C) = 0.11 / 0.25 = 0.44, making statement B true. The probability of both events is 0.11, and intersection is commutative, so P(C ∩ D) = P(D ∩ C). The two conditional probabilities are not equal.

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