Question

A survey 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)=

Answer 1

Explanation:

Cats are better (sorry, I just had to say it.)

Answer 2

I) P(cat│dog) =
`(0.15)/(0.38)`

II) These events are not independent

III) P(cat or dog)= 0.7

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 ) =
`P(cat\cap dog)` = 0.15

solution :

i) By formula P(A│B) =
`(P(A\cap B))/(P(B))`

P(cat│dog)=
`(P(cat\cap dog))/(P(dog))`

P(cat│dog) =
`(0.15)/(0.38)`

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(A\cup B)= P(A) + P(B)-P(A\cap B)`

P(cat or dog) =
`P(cat\cup dog)= P(cat) + P(dog)-P(cat\cap dog)`

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

P(cat or dog)= 0.7