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 :
(1)
Similarly ,
![P(C\cap D)=P(C|D)\times P(D)\\\\=0.22\times0.36=0.0792](https://tex.z-dn.net/?f=P%28C%5Ccap%20D%29%3DP%28C%7CD%29%5Ctimes%20P%28D%29%5C%5C%5C%5C%3D0.22%5Ctimes0.36%3D0.0792)
Hence, the probability that a randomly selected family owns both a dog and a cat : 0.0792
b) Again, using (2)
![P(D|C)=\dfrac{P(C\cap D)}{P(C)}\\\\=\dfrac{0.0792}{0.30}=0.264](https://tex.z-dn.net/?f=P%28D%7CC%29%3D%5Cdfrac%7BP%28C%5Ccap%20D%29%7D%7BP%28C%29%7D%5C%5C%5C%5C%3D%5Cdfrac%7B0.0792%7D%7B0.30%7D%3D0.264)
Hence, the conditional probability that a randomly selected family owns a dog given that it owns a cat = 0.264