Answer:
I) P(cat│dog) = [tex]\frac{0.15}{0.38}[/tex]
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 ) = [tex]P(cat\cap dog)[/tex] = 0.15
solution :
i) By formula P(A│B) =[tex]\frac{P(A\cap B)}{P(B)}[/tex]
P(cat│dog)= [tex]\frac{P(cat\cap dog)}{P(dog)}[/tex]
P(cat│dog) = [tex]\frac{0.15}{0.38}[/tex]
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 : [tex]P(A\cup B)= P(A) + P(B)-P(A\cap B)[/tex]
P(cat or dog) = [tex]P(cat\cup dog)= P(cat) + P(dog)-P(cat\cap dog)[/tex]
P(cat or dog)= 0.47 + 0.38 - 0.15
P(cat or dog)= 0.7
