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.
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)

Answer 1

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

Answer 2

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