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Why do I subtract .2 twice

schinb65

New member
Jan 1, 2013
12
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.

I am Told that the answer is .5, I did this problem 2 different ways and I received different answers, and .5 is one of the answers.

If I draw a Venn Diagram I receive .5 and that makes sense.
I also tried this, \(P[A\cup B]=P[A]+P-P[A \cap B]\),
I assume that I have something incorrect in the formula since I do not get the right answer. Would I be able to use this formula?
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
Welcome to MHB, schinb65!

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.

I am Told that the answer is .5, I did this problem 2 different ways and I received different answers, and .5 is one of the answers.

If I draw a Venn Diagram I receive .5 and that makes sense.
I also tried this, \(P[A\cup B]=P[A]+P-P[A \cap B]\),
I assume that I have something incorrect in the formula since I do not get the right answer. Would I be able to use this formula?


A Venn Diagram is best. ;)

But yes, you can also use that formula.
\begin{aligned}
P[\text{A or B but not both}] &= P[A\cup B]-P[A \cap B] \\
&=\big(P[A]+P-P[A \cap B]\big)-P[A \cap B] \\
&=P[A]+P-2P[A \cap B] \\
&=60\%+30\%-2\times 20\% \\
&=50\% \\
\end{aligned}