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)

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

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