Question

In a certain town 30% families own a scooter and 40% own a car, 50% own neither a scooter nor a car, 2000 families own both scooter and car. Consider following statements in this regards

1. 20% of families own both scooter and car
2. 35% of families own either a car or a scooter
3. 10000 families live in town
   Which of the above statements are correct?
   (A) 2 and 3
   (B) 1, 2 and 3
   (C) 1 and 2
   (D) 1 and 3

Answer 1

In this type of question we have to use the concept of probability. Here, we will consider the total number of families in the town is equal to x and the percentages given to us we have to consider that as the probabilities of the different events. We have to use $p+q=1$ where $p$ and $q$ are the probabilities of success and failure respectively. Also we have to use the formula of probability of union of two sets i.e. $P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right)$

Complete step-by-step solution:
Now here, we have to choose the correct statements with the help of given information.
We have given that, in a certain town 30% families own a scooter and 40% own a car, 50% own neither a scooter nor a car and 2000 families own both scooter and car.
Let us assume that $x$ represents the total number of families in the town.
Suppose A denotes families who own scooter and B denotes families who own car

& \Rightarrow P\left( A \right)=30\% \\\ & \Rightarrow P\left( B \right)=40\% \\\ & \Rightarrow P\left( A\bigcup B \right)'=50\% \\\ \end{aligned}$$ Now as we know that, if $$p$$ and $$q$$ are the probabilities of success and failure respectively then $$p+q=1$$ $$\begin{aligned} & \Rightarrow P\left( A\bigcup B \right)=1-P\left( A\bigcup B \right)' \\\ & \Rightarrow P\left( A\bigcup B \right)=1-50\% \\\ & \Rightarrow P\left( A\bigcup B \right)=1-\dfrac{50}{100} \\\ & \Rightarrow P\left( A\bigcup B \right)=50\% \\\ \end{aligned}$$ This means 50% of families own either a car or a scooter. Now by using the probability of union of two events we have, $$\begin{aligned} & \Rightarrow P\left( A\bigcup B \right)=P\left( A \right)+P\left( B \right)-P\left( A\bigcap B \right) \\\ & \Rightarrow 50\%=30\%+40\%-P\left( A\bigcap B \right) \\\ & \Rightarrow P\left( A\bigcap B \right)=30\%+40\%-50\% \\\ & \Rightarrow P\left( A\bigcap B \right)=20\% \\\ \end{aligned}$$ This means that in that town 20% of families own both a scooter and car. But we have given that 2000 families own both scooter and car and the total number of families is equal to $$x$$ $$\begin{aligned} & \Rightarrow 20\%\text{ of }x=2000 \\\ & \Rightarrow \dfrac{20}{100}\times x=2000 \\\ & \Rightarrow 0.2x=2000 \\\ & \Rightarrow x=\dfrac{2000}{0.2} \\\ & \Rightarrow x=10000 \\\ \end{aligned}$$ Hence, there are 10000 families in that town. Therefore we can observe that the first and third statements are correct. **Thus, option (D) is the correct option.** **Note:** In this type of question students have to take care in representing probability for neither scooter nor car. Also students have to remember that when they calculate the probability of either scooter or car at that time they have to subtract the probability of neither scooter nor car from 1 as the total probability is always equal to 1.