# Probability a person chosen at random owns an automobile or a house, but not both

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 $\Pr(A~\text{and}~B) = 2 \cdot 0.2$?

• You computed the probability of "owning an automobile or a house." They asked for the probability of "owning an automobile or a house but not both." – angryavian May 30 '18 at 6:16
• @angryavian I still don't get it, I thought "owning an automobile or a house" is the same as "owning an automobile or a house but not both" ?? – pino231 May 30 '18 at 6:20
• Please type your question rather than posting an image. Images cannot be searched. – N. F. Taussig May 30 '18 at 8:54
• @pino231, in fact "owning an automobile or a house" in this case means "owning an automobile, or a house, or both". This usage of the word "or" -- the so-called "inclusive or" -- is standard in mathematics. If we want the opposite (the exclusive or) it is almost always explicitly stated, as it is in your question. – Mees de Vries May 30 '18 at 9:26

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.

• ahh so the question is actually asking the probability that a person chosen at random owns an automobile or a house, OR not both – pino231 Jun 1 '18 at 3:20
• Not quite. The word but should be read as and. The question is asking the probability that a person chosen at random owns an automobile or a house and not both: $\Pr(A~\triangle~B) = \Pr(A \cup B) - \Pr(A \cap B)$. – N. F. Taussig Jun 1 '18 at 9:17

There's no ordering in this question. Using the addition rule is fine, but if you want to use it, note that:

$\mathbb{P}(A) + \mathbb{P}(B) = \Big( \mathbb{P}(A \setminus B) + \mathbb{P}(A \cap B) \Big) + \Big( \mathbb{P}(B \setminus A) + \mathbb{P}(A \cap B) \Big)$.

This should probably help.

10 $\%$ of people don't have a house or automobile so you can leave them. There is 90 $\%$ of people left, also 20$\%$ have both a house and a automobile. This 20% of people is included in the 60% and 30%, so you have to substract this percentage. What you get is 40% + 10% = 50%. So you were probably right by assuming that you have to count the last term $P(A \cap B)$ two times.

• If we subtract $\Pr(A \cup B)$ two times, we will obtain a negative answer. Instead, we must subtract $\Pr(A \cap B)$ twice. – N. F. Taussig May 30 '18 at 9:13
• Yes of course, that's what I meant. Will edit! – WarreG May 30 '18 at 9:37