Question

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident are 0.4, 0.03 and 0.15 respectively. One of the insured person meets with an accident. What is the probability that he is a scooter driver?

Answer 1

Hint: We will use the Bayes theorem of probability to solve the above question. Bayes theorem is a way to find a probability when we know certain other probabilities.
Given by as follows:
$P\left( {A/B} \right) = \dfrac{{P\left( A \right)\,\,\,P\left( {B/A} \right)}}{{P\left( B \right)}}$
Where, ${\rm{P}}\left( {{\rm{A}}/{\rm{B}}} \right) =$Probability of A when B happens.
${\rm{P}}\left( {{\rm{B}}/{\rm{A}}} \right) =$Probability of B when A happens.
${\rm{P}}\left( {\rm{A}} \right) =$Probability of event A.
${\rm{P}}\left( {\rm{B}} \right) =$Probability of event B.

Complete step-by-step solution -
We have been given that an insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers and the probability of an accident are 0.4, 0.03 and 0.15 respectively. If one of the insured persons meets with an accident, then we have to find what is the probability that he is a scooter driver.
Let us write down the data we have as below,
Total number of drivers $= 2000 + 4000 + 6000$
$= 12000$
Let us consider that ${\rm{P}}\left( {\rm{A}} \right) =$ Probability of scooter driver $= \dfrac{{{\text{Number of Scooter driver}}}}{{{\text{ Total number of drivers}}}}$
$\begin{array}{l} = \dfrac{{2000}}{{12000}}\\\ = \dfrac{1}{6}\end{array}$
Next, let us consider that ${\rm{P}}\left( {\rm{B}} \right) =$Probability of car driver $= \dfrac{{{\text{Number of Car driver}}}}{{{\text{Total number of drivers}}}}$
$\begin{array}{l} = \dfrac{{4000}}{{12000}}\\\ = \dfrac{1}{3}\end{array}$
Now, we will consider that ${\rm{P}}\left( {\rm{C}} \right) =$Probability of truck driver $= \dfrac{{{\text{Number of Truck driver}}}}{{{\text{Total number of drivers}}}}$
$\begin{array}{l} = \dfrac{{6000}}{{12000}}\\\ = \dfrac{1}{2}\end{array}$
Let E be the event that the person meets with an accident. Then, we have been given,
$\begin{array}{l}{\rm{P}}\left( {{\rm{E}}/{\rm{A}}} \right) = 0.01 = \dfrac{1}{{100}}\\\\{\rm{P}}\left( {{\rm{E}}/{\rm{B}}} \right) = 0.03 = \dfrac{3}{{100}}\\\\{\rm{P}}\left( {{\rm{E}}/C} \right) = 0.15 = \dfrac{{15}}{{100}}\end{array}$

Now, we have to find the probability of scooter driver if the accident took place i.e.${\rm{P}}\left( {{\rm{A}}/{\rm{E}}} \right)$.
We will have the Bayes theorem to find this value and the formula for the Bayes theorem is given by,
$P\left( {A/E} \right) = \dfrac{{P\left( A \right).\,P\left( {E/A} \right)}}{{P\left( A \right)\,.\,P\left( {E/A} \right) + P\left( B \right)\,.\,P\left( {E/B} \right) + P\left( C \right)\,.\,P\left( {E/C} \right)}}$
Substituting all the values, we get
$= \dfrac{{\dfrac{1}{6} \times \dfrac{1}{{100}}}}{{\dfrac{1}{6} \times \dfrac{1}{{100}} + \dfrac{1}{3} \times \dfrac{3}{{100}} + \dfrac{1}{2} \times \dfrac{{15}}{{100}}}}$
Taking $\dfrac{1}{{100}}$ as common from numerator and denominator and cancelling them, we get
$\begin{array}{l} = \dfrac{{\dfrac{1}{6}}}{{\dfrac{1}{6} + 1 + \dfrac{{15}}{2}}}\\\ = \dfrac{{\dfrac{1}{6}}}{{\dfrac{{1 + 6 + 45}}{6}}}\\\ \Rightarrow P\left( {A/E} \right) = \dfrac{1}{{52}}\end{array}$
Therefore, the required probability is equal to $\dfrac{1}{{52}}$.

Note: The main problem of probability questions is that we are unable to think which rule of probability is applicable to the given question. So, remember that if you have been given the other probabilities value of events and they asked to find the probability if any event has already happened then you must use the Bayes formula.