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Question: In a certain town, \[60\% \] of the families own a car, \[30\% \] own a house and \[20\% \] own both...

In a certain town, 60%60\% of the families own a car, 30%30\% own a house and 20%20\% own both car and house. If a family is randomly chosen, then what is the probability that this family owns a car or a house but not both?
A.0.50.5
B.0.70.7
C.0.10.1
D.0.90.9

Explanation

Solution

Here, we will find the probability that the family owns a car or a house but not both. We will find the probability of the families owning at least a car or a house by using the OR rule. We will use the probability in symmetric difference to find the probability that the family owns a car or a house but not both. Thus the required probability.

Formula Used:
We will use the following formula:
1.The probability is given by the formula P(A)=n(A)n(S)P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}} where n(A)n\left( A \right) is the number of favorable outcomes and n(S)n\left( S \right) is the total number of outcomes.
2.OR rule: P(A∪B)=P(A)+P(B)−P(A∩B)P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right)
3.P(AΔB)=P(A∪B)−P(A∩B)P\left( {A\Delta B} \right) = P\left( {A \cup B} \right) - P\left( {A \cap B} \right)

Complete step-by-step answer:
We are given that in a certain town, 60%60\% of the families own a car, 30%30\% own a house and 20%20\% own both car and house.
Since the number is given in Percentage, the total number is 100100.
So, the sample space would be 100%100\%
Let AA be the event of choosing a family owning a car and BB be the event of choosing a family owning a house.
The probability is given by the formula P(A)=n(A)n(S)P\left( A \right) = \dfrac{{n\left( A \right)}}{{n\left( S \right)}}
⇒\Rightarrow Probability of choosing a family owning a carP(A)=60100P\left( A \right) = \dfrac{{60}}{{100}}
The probability is given by the formula P(B)=n(B)n(S)P\left( B \right) = \dfrac{{n\left( B \right)}}{{n\left( S \right)}}
⇒\Rightarrow Probability of choosing a family owning a houseP(B)=30100P\left( B \right) = \dfrac{{30}}{{100}}
The probability is given by the formula P(A∩B)=n(A∩B)n(S)P\left( {A \cap B} \right) = \dfrac{{n\left( {A \cap B} \right)}}{{n\left( S \right)}}
⇒\Rightarrow Probability of choosing a family owning a car and a houseP(A∩B)=20100P\left( {A \cap B} \right) = \dfrac{{20}}{{100}}
Now, we will use the OR rule in Probability, to find the probability of the families owning at least a car or a house.
Now, by using the OR rule in probability P(A∪B)=P(A)+P(B)−P(A∩B)P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) and substituting the values, we get
⇒P(A∪B)=60100+30100−20100\Rightarrow P\left( {A \cup B} \right) = \dfrac{{60}}{{100}} + \dfrac{{30}}{{100}} - \dfrac{{20}}{{100}}
⇒P(A∪B)=70100\Rightarrow P\left( {A \cup B} \right) = \dfrac{{70}}{{100}}
Now, we will find the probability that the family owns a car or a house but not both car and house.
⇒P(AΔB)=P(A∪B)−P(A∩B)\Rightarrow P\left( {A\Delta B} \right) = P\left( {A \cup B} \right) - P\left( {A \cap B} \right)
⇒P(AΔB)=70100−20100\Rightarrow P\left( {A\Delta B} \right) = \dfrac{{70}}{{100}} - \dfrac{{20}}{{100}}
⇒P(AΔB)=50100\Rightarrow P\left( {A\Delta B} \right) = \dfrac{{50}}{{100}}
⇒P(AΔB)=0.5\Rightarrow P\left( {A\Delta B} \right) = 0.5
Therefore, the probability that the family owns a car or a house but not both car and house is 0.50.5
Option (A) is the correct answer.

Note: We know that the OR rule in probability states that the outcome has to satisfy one condition or other condition or both the conditions at the same time. If the event is a mutually exclusive event, then the probabilities of one condition and the other condition has to be added, so the event is also called a disjoint event. AND rule in probability states that that the outcome has to satisfy both the conditions at the same time. We should always remember that if the number is in percentage then the total will always be 100.