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4 December, 07:04

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.

1. P (C | D) = 0.78

2. P (D | C) = 0.44

3. P (C ∩ D) = 0.11

4. P (C ∩ D) = P (D ∩ C)

5. P (C | D) = P (D | C)

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Answers (2)
  1. 4 December, 07:07
    0
    P (D | C) = 0.44 P (C ∩ D) = 0.11 P (C ∩ D) = P (D ∩ C)
  2. 4 December, 07:16
    0
    Check the venn diagram of the problem.

    let the total number of the clients be 100.

    since there are 11 clients with both cats and dogs,

    there are 32-11=21 clients with dogs but not cats

    and 25-11=14 clients with cat but not dogs.

    The number of clients with neither dogs nor cats is given by:

    100 - (21+11+14) = 100-46=54

    so we have the following:

    P (C) = 25/100=0.25

    P (D) = 32/100=0.32

    P (C∩D) = P (D∩C) = 11/100=0.11

    we also have the formula P (A|B) = P (A∩B) / P (B), the formula of conditional probability

    now we check each choice:

    1. P (C|D) = P (C∩D) / P (D) = 0.11/0.32=0.34

    2. P (D|C) = P (D∩C) / P (C) = 0.11/0.25=0.44

    3 and 4 are already checked, and they are true

    5. by 1 and 2, not true

    Answer: 2, 3, 4 are True
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