Answer: a) 0.0792 b) 0.264
Step-by-step explanation:
Let Event D = Families own a dog .
Event C = families own a cat .
Given : Probability that families own a dog : P(D)=0.36
Probability that families own a dog also own a cat : P(C|D)=0.22
Probability that families own a cat : P(C)= 0.30
a) Formula to find conditional probability :
[tex]P(B|A)=\dfrac{P(A\cap B)}{P(A)}\\\\\Rightarrow P(A\cap B)=P(B|A)\times P(A) [/tex] (1)
Similarly ,
[tex]P(C\cap D)=P(C|D)\times P(D)\\\\=0.22\times0.36=0.0792[/tex]
Hence, the probability that a randomly selected family owns both a dog and a cat : 0.0792
b) Again, using (2)
[tex]P(D|C)=\dfrac{P(C\cap D)}{P(C)}\\\\=\dfrac{0.0792}{0.30}=0.264[/tex]
Hence, the conditional probability that a randomly selected family owns a dog given that it owns a cat = 0.264