Question

A survey of households revealed that 38% have a dog, 47% have a cat, and 15% have both a cat and a dog.

Answer 1

Answer:

Step-by-step explanation:

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

Answer 2

Answer:

I) P(cat│dog) = $\frac{0.15}{0.38}$

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

solution :

i) By formula P(A│B) =$\frac{P(A\cap B)}{P(B)}$

P(cat│dog)= $\frac{P(cat\cap dog)}{P(dog)}$

P(cat│dog) = $\frac{0.15}{0.38}$

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