Asurvey of households revealed that 38% have a dog, 47% have a cat, and 15% have both a cat and a dog. given that a household owns a dog, what is the probability that it also owns a cat? p(cat│dog)= since p(cat│dog)=39.47% and p(cat)=47%, are the events independent or not independent? what is the probability of a household owning a cat or a dog? p(cat or dog)=

I) P(cat│dog) =

II) These events are not independent

III) P(cat or dog)=  0.7

Step-by-step explanation:

Given : Households have  dogs = 38%

           So, P(dog) = 0.38

           Households have  cats = 47%

           So, P(cats) = 0.47

             Households have both dogs and cats  = 15%

           So, P(both dog and cat ) = = 0.15

solution :

i) By formula P(A│B) =

P(cat│dog)=

P(cat│dog) =

ii)  P(cat│dog)=39.47% = 0.39 and P(cat)=47% = 0.47, are the events not independent

Because condition for independent events in conditional probability is P(A|B)=P(A)

but P(cat│dog) ≠P(cat) i.e. 0.39≠0.47

So, these events are not independent

iii) P(cat or dog) = ?

"or" means union

Formula :

P(cat or dog) =

P(cat or dog)= 0.47 + 0.38 - 0.15

P(cat or dog)=  0.7

could u provide us some answer choices or hmm idk maybe a table for us to anser with ? ?