Question

An insurance company insured 3000 scooters, 4000 cars and 5000 trucks. The probabilities of the accident involving a scooter, a car and a truck are 0.02, 0.03 and 0.04 respectively. One of the insured vehicles meet with an accident. Find the probability that it is a (i) scooter (ii) car (iii) truck.

Answer 1

Let E₁, E₂ and E₃ denote the events that the vehicle is a scooter, a car and a truck, respectively.

Let A be the event that the vehicle meets with an accident.

It is given that there are 3000 scooters, 4000 cars and 5000 trucks.

Total number of vehicles = 3000 + 4000 + 5000 = 12000

P(E₁) = $\frac{3000}{12000}=\frac{1}{4}$

P(E₂) =$\frac{4000}{12000}=\frac{1}{3}$

P(E₃) = $\frac{5000}{12000}=\frac{5}{12}$

The probability that the vehicle, which meets with an accident, is a scooter is given by P (E₁/A).

$$\begin{gathered} \mathrm{Now}, \\ P\left(A/E_1\right)=0.02=\frac{2}{100} \\ P\left(A/E_2\right)=0.03=\frac{3}{100} \\ P\left(A/E_3\right)=0.04=\frac{4}{100} \\ \mathrm{Using}\mathrm{Bayes}\text{'}\mathrm{theorem},\mathrm{we}\mathrm{get} \\ \left(\mathrm{i}\right)\mathrm{Required}\mathrm{probability}=P\left(E_1/A\right)=\frac{P\left(E_1\right)P\left(A/E_1\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)++P\left(E_2\right)P\left(A/E_2\right)} \\ =\frac{{\displaystyle \frac{1}{4}}\times {\displaystyle \frac{2}{100}}}{{\displaystyle \frac{1}{4}}\times {\displaystyle \frac{2}{100}}+{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{3}{100}}+{\displaystyle \frac{5}{12}}\times {\displaystyle \frac{4}{100}}} \\ =\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}+1+{\displaystyle \frac{5}{3}}}=\frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{3+6+10}{6}}}=\frac{3}{19} \\ \left(\mathrm{ii}\right)\mathrm{Required}\mathrm{probability}=P\left(E_2/A\right)=\frac{P\left(E_2\right)P\left(A/E_2\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)++P\left(E_2\right)P\left(A/E_2\right)} \\ =\frac{{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{3}{100}}}{{\displaystyle \frac{1}{4}}\times {\displaystyle \frac{2}{100}}+{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{3}{100}}+{\displaystyle \frac{5}{12}}\times {\displaystyle \frac{4}{100}}} \\ =\frac{{\displaystyle 1}}{{\displaystyle \frac{1}{2}}+1+{\displaystyle \frac{5}{3}}}=\frac{1}{{\displaystyle \frac{3+6+10}{6}}}=\frac{6}{19} \\ \left(\mathrm{iii}\right)\mathrm{Required}\mathrm{probability}=P\left(E_2/A\right)=\frac{P\left(E_3\right)P\left(A/E_3\right)}{P\left(E_1\right)P\left(A/E_1\right)+P\left(E_2\right)P\left(A/E_2\right)++P\left(E_2\right)P\left(A/E_2\right)} \\ =\frac{{\displaystyle \frac{5}{12}}\times {\displaystyle \frac{4}{100}}}{{\displaystyle \frac{1}{4}}\times {\displaystyle \frac{2}{100}}+{\displaystyle \frac{1}{3}}\times {\displaystyle \frac{3}{100}}+{\displaystyle \frac{5}{12}}\times {\displaystyle \frac{4}{100}}} \\ =\frac{{\displaystyle \frac{5}{3}}}{{\displaystyle \frac{1}{2}}+1+{\displaystyle \frac{5}{3}}}=\frac{{\displaystyle \frac{5}{3}}}{{\displaystyle \frac{3+6+10}{6}}}=\frac{10}{19} \end{gathered}$$