Aveterinarian 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)

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

2, 3, 4 are True

Ithink the answer is c just trying to others