Suppose $36\%$ of families own a dog, $30\%$ of families own a cat, and $22\%$ of the families that have a dog also have a cat. A family is chosen at random and found to have a cat. What is the probability they also own a dog?
D: Family owns a dog.
C: Family owns a cat.
"$22\%$ of the families that have a dog also have a cat" - why this is not $D \cap C$ ?
On
In the question the phrase that the family is selected at random is found to have a cat implies that our sample space is now reduced to those containing cats as sample space N(C),as the event has already occured and we have cat in our hand and now probability that in these selected families(having cats) the family that also contains dog is P(D when C has been found out)=N(C int D) / N(C) now divide num and den by total sample space this results in conditional probability expression I have just elaborated that if u do P(C int D) then you are not considering the fact that the selected family is already having cat Hope that it helps! Please vote if you like the answer. int:- intersection
"$22\%$ of the families that have a dog also have a cat"
This is a fraction of the families that have a dog, rather than of all the families.
Thus it is a conditional probability, $\mathsf P(C\mid D)$, rather than the conjunct probability, $\mathsf P(C\cap D)$.
Hence, you seek the fraction of families that have a cat which also have a dog. $\mathsf P(D\mid C)$.
And you should know that by the definition of conditional probability $$\mathsf P(C\cap D)=\mathsf P(D)~\mathsf P(C\mid D)~=~\mathsf P(C)~\mathsf P(D\mid C)$$