question_answer

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident involving a scooter, a car and a truck are 0.01, 0.03 and 0.15, respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?      

Answer:

Let us define the following events: \[{{E}_{1}}=\] Insured person is a scooter driver \[{{E}_{2}}=\] Insured person is a car driver \[{{E}_{3}}=\] Insured person is a truck driver A = Insured person meets with an accident Here, total number of insured vehicles \[=2000+4000+6000=12000\] \[\therefore \] \[P({{E}_{1}})=\frac{2000}{12000}=\frac{1}{6},\] \[P({{E}_{2}})=\frac{2000}{12000}=\frac{1}{3}\] and \[P({{E}_{3}})=\frac{6000}{12000}=\frac{1}{2}\] Also given, \[P\left( \frac{A}{{{E}_{1}}} \right)=\] Probability that scooter driver meets with an accident = 0.01 \[P\left( \frac{A}{{{E}_{2}}} \right)=\] Probability that car driver meets with an accident = 0.03 \[P\left( \frac{A}{{{E}_{3}}} \right)=\] Probability that truck driver meets with an accident = 0.15                                Now, P (person meets with an accident is a scooter driver),                                          [by Baye's theorem] On putting all the above values, we get \[P\left( \frac{{{E}_{1}}}{A} \right)=\frac{\frac{1}{6}\times 0.01}{\left( \frac{1}{6}\times 0.01 \right)+\left( \frac{1}{3}\times 0.03 \right)+\left( \frac{1}{2}\times 0.15 \right)}\] \[=\frac{\frac{1}{6}\times \frac{1}{100}}{\left( \frac{1}{6}\times \frac{1}{100} \right)+\left( \frac{1}{3}\times \frac{3}{100} \right)+\left( \frac{1}{2}\times \frac{15}{100} \right)}\] \[=\frac{\frac{1}{6}}{\frac{1}{6}+1+\frac{15}{2}}=\frac{\frac{1}{6}}{\frac{1+6+45}{6}}=\frac{1}{52}\]