A marketing survey indicates that $60\%$ of the population owns an automobile, $30\%$ owns a house, and $20\%$ owns both an automobile and a house. Calculate the probability that a person chosen at random owns an automobile or a house, but not both.
(A) $0.4$
(B) $0.5$
(C) $0.6$
(D) $0.7$
(E) $0.9$
So the answer given for this question is $0.5$, but it doesn't make sense to me from the perspective of addition rule in probability. If we know that $\Pr(\text{Owning Automobile}) = 0.6$ and the $\Pr(\text{Owning a House}) = 0.3$ and the $\Pr(\text{Owning both an automobile and a house}) = 0.2$,
according to the formula probability of $\Pr(A)~\text{or}~\Pr(B) = \Pr(A) + \Pr(B) – \Pr(A~\text{and}~B)$, wouldn't the answer be $0.6+0.3-0.2=0.7$?
Is it because the ordering matters? so the last term $\Pr(A~\text{and}~B) = 2 \cdot 0.2$?
Best Answer
Let event $A$ be owning an automobile. Let event $B$ be owning a house.
Consider the diagram below:
If we simply add the probabilities that a person owns an automobile and a person owns a house, we will have added the probability that a person owns both an automobile and a house twice. Thus, to find the probability that a person owns an automobile or a house, we must subtract the probability that a person owns both an automobile and a house from the sum of the probabilities that a person owns an automobile and that a person owns a house.
$$\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B)$$
Note that on the left-hand side of your equation, you should have written $\Pr(A \cup B)$ or $\Pr(A~\text{or}~B)$ rather than $\Pr(A)~\text{or}~\Pr(B)$.
Since we are given that $60\%$ of the population owns an automobile, $30\%$ of the populations owns a house, and $20\%$ owns both, $\Pr(A) = 0.60$, $\Pr(B) = 0.30$, and $\Pr(A \cap B) = 0.20$. Hence, the probability that a person owns an automobile or a house is $$\Pr(A \cup B) = \Pr(A) + \Pr(B) - \Pr(A \cap B) = 0.60 + 0.30 - 0.20 = 0.70$$ However, the question asks for the probability that a person owns an automobile or a house, but not both. That means we must subtract the probability that a person owns an automobile and a house from the probability that the person owns an automobile or a house.
$$\Pr(A~\triangle~B) = \Pr(A \cup B) - \Pr(A \cap B) = 0.70 - 0.20 = 0.50$$
In terms of the Venn diagram, $A \cup B$ is the region enclosed by the two circles, while $A~\triangle~B = (A \cup B) - (A \cap B) = (A - B) \cup (B - A)$ is the region enclosed by the two circles except the region where the sets intersect.
Since $A - B = A - (A \cap B)$, $$\Pr(A - B) = \Pr(A) - \Pr(A \cap B) = 0.60 - 0.20 = 0.40$$
Since $B - A = B - (A \cap B)$, $$\Pr(B - A) = \Pr(B) - \Pr(A \cap B) = 0.30 - 0.20 = 0.10$$
Hence,
$$\Pr(A~\triangle~B) = \Pr(A - B) + \Pr(B - A) = 0.40 + 0.10 = 0.50$$ which agrees with the result obtained above.