In a certain town,60 % of the families own a | vedclass

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(A) Let $A$ be the set of families who own a car and $B$ be the set of families who own a house.Given: $P(A) = 0.60$,$P(B) = 0.30$,and $P(A \cap B) =

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$0.5$

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(A) Let $A$ be the set of families who own a car and $B$ be the set of families who own a house.Given: $P(A) = 0.60$,$P(B) = 0.30$,and $P(A \cap B) = 0.20$.We need to find the probability that a family owns a car or a house but not both,which is represented by the symmetric difference $P(A \Delta B)$.The formula for symmetric difference is $P(A \Delta B) = P(A \cup B) - P(A \cap B)$.First,calculate $P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.60 + 0.30 - 0.20 = 0.70$.Now,$P(A \Delta B) = P(A \cup B) - P(A \cap B) = 0.70 - 0.20 = 0.50$.Thus,the probability is $0.5$.

In a certain town,$60 \%$ of the families own a car,$30 \%$ own a house and $20 \%$ own both a car and a house. If a family is randomly chosen,what is the probability that this family owns a car or a house but not both?

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