Answer:
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
4.50 and 8 I think the answer is
Answer: 3 is the missing numerator.
Step-by-step explanation:
4/6 + n/6 = 1 1/6
n = 3
4/6 + 3/6 = 1 1/6
Answer: Nick needs to sell 20 treats to break even.