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Worked Examples · Example 17

Q.In a certain town, 25% of the families own a phone, 15% own a car, and 65% families own neither a phone nor a car. 2000 families own both a car and a phone. Find how much percentage of families own either a car or a phone. Also, find how many families live in the town.

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Get the "either" percentage from 100%−neither100\%-\text{neither}, then find the "both" percentage via the union formula, then scale using the given count of 2000 families.

P(Phone∪Car)=100%−P(neither)P(\text{Phone}\cup\text{Car})=100\%-P(\text{neither}); P(Phone∩Car)=P(Phone)+P(Car)−P(Phone∪Car)P(\text{Phone}\cap\text{Car})=P(\text{Phone})+P(\text{Car})-P(\text{Phone}\cup\text{Car}); Total families=count owning bothP(both)\text{Total families}=\dfrac{\text{count owning both}}{P(\text{both})}.

  1. Given: P(Phone)=25%P(\text{Phone})=25\%, P(Car)=15%P(\text{Car})=15\%, P(neither)=65%P(\text{neither})=65\%, families owning both =2000=2000.
  2. Percentage owning either (union): P(Phone∪Car)=100%−65%=35%P(\text{Phone}\cup\text{Car})=100\%-65\%=35\%.
  3. Percentage owning both: P(Phone∩Car)=P(Phone)+P(Car)−P(Phone∪Car)=25%+15%−35%=40%−35%=5%P(\text{Phone}\cap\text{Car})=P(\text{Phone})+P(\text{Car})-P(\text{Phone}\cup\text{Car})=25\%+15\%-35\%=40\%-35\%=5\%.
  4. Since 5%5\% of the total families =2000=2000 families: …

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