Question

In a certain community, 36 percent of the families own a dog and 22 percent of the families that own a dog also own a cat. In addition, 30 percent of the families own a cat. What is (a) the probability that a randomly selected family owns both a dog and a cat? (b) the conditional probability that a randomly selected family owns a dog given that it owns a cat?

Answer 1

To find the probability that a randomly selected family owns both a dog and a cat, multiply the probabilities of owning a dog and owning a cat given that a family owns a dog. The probability is 7.92%. The conditional probability that a randomly selected family owns a dog given that it owns a cat is 26.4%.

Step-by-step explanation:

To find the probability that a randomly selected family owns both a dog and a cat, we need to multiply the probabilities of owning a dog and owning a cat given that a family owns a dog. Let's assume there are 100 families in the community:

Number of families that own a dog = 36% of 100 = 36

Number of families that own a cat = 30% of 100 = 30

Number of families that own both a dog and a cat = 22% of 36 = 0.22 * 36 = 7.92

Therefore, the probability that a randomly selected family owns both a dog and a cat is 7.92/100 = 0.0792 or 7.92%

(b) To find the conditional probability that a randomly selected family owns a dog given that it owns a cat, we can use the formula:

P(Dog|Cat) = P(Dog and Cat) / P(Cat)

Substituting the values:

P(Dog|Cat) = 0.0792 / 0.30 = 0.264 or 26.4%

Answer 2

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 :

`P(B|A)=(P(A\cap B))/(P(A))\\\\\Rightarrow P(A\cap B)=P(B|A)* P(A)` (1)

Similarly ,

`P(C\cap D)=P(C|D)* P(D)\\\\=0.22*0.36=0.0792`

Hence, the probability that a randomly selected family owns both a dog and a cat : 0.0792

b) Again, using (2)

`P(D|C)=(P(C\cap D))/(P(C))\\\\=(0.0792)/(0.30)=0.264`

Hence, the conditional probability that a randomly selected family owns a dog given that it owns a cat = 0.264