Question

An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probabilities of an accident involving a scooter driver, car driver and truck driver are 0.01, 0.03 and 0.15 respectively. One of the insured persons meets with an accident. The probability that person is a scooter driver is
(a) $\dfrac{1}{52}$
(b) $\dfrac{3}{52}$
(c) $\dfrac{15}{52}$
(d) $\dfrac{19}{52}$

Answer 1

Hint: Here, we will use the concept of conditional probability to find the probability that the person who meets with an accident is a scooter driver. If A and B are two events in a sample space S, then the conditional probability of A given B is defined as: $P\left( \dfrac{A}{B} \right)=\dfrac{P\left( A\cap B \right)}{P\left( B \right)}...........\left( 1 \right)$

Complete step-by-step answer:
Conditional probability is a measure of the probability of an event occurring given that another event has (by assumption, presumption, assertion or evidence ) occurred. If the event of interest is A and the event B is known or assumed to have occurred, “ the conditional probability of A given B”, or, “ the probability of A under the condition B” is usually written as $P\left( \dfrac{A}{B} \right)$ .
Let us consider that ‘S’ be the event that the driver is a scooter driver.
‘C’ be the event that driver be the car driver.
‘T’ be the event that the driver is a truck driver.
‘A’ be the event that an accident took place.
We have to find the probability of an accident with a scooter driver. It means that we have to find the value of $P\left( \dfrac{A}{B} \right)$ .
Using equation (1), we can write:
$P\left( \dfrac{S}{A} \right)=\dfrac{P\left( S\cap A \right)}{P\left( A \right)}............\left( 2 \right)$
It is also given that $P\left( \dfrac{A}{S} \right)=0.01$ , $P\left( \dfrac{A}{C} \right)=0.03$ and $P\left( \dfrac{A}{T} \right)=0.15$.
Also we have, $P\left( \dfrac{A}{S} \right)=\dfrac{P\left( A\cap S \right)}{P\left( S \right)}$
So, $P\left( A\cap S \right)=P\left( \dfrac{A}{S} \right)\times P\left( S \right)$
Therefore, $P\left( S\cap A \right)=P\left( \dfrac{A}{S} \right)\times P\left( S \right)$
Putting this value in equation (1), we get:
$P\left( \dfrac{S}{A} \right)=\dfrac{P\left( \dfrac{A}{S} \right)\times P\left( S \right)}{P\left( A \right)}............\left( 3 \right)$
We have been given that $P\left( \dfrac{A}{S} \right)=0.01$.
And, $P\left( S \right)=\dfrac{\text{total number of scooter drivers}}{\text{total number of drivers}}=\dfrac{2000}{2000+4000+6000}=\dfrac{2000}{12000}=\dfrac{1}{6}$
Therefore, we have:
 $\begin{align}
  & P\left( A \right)=P\left( S \right)\times P\left( \dfrac{A}{S} \right)+P\left( C \right)\times P\left( \dfrac{A}{C} \right)+P\left( T \right)\times P\left( \dfrac{A}{T} \right) \\
 & \Rightarrow P\left( A \right)=\dfrac{2000}{12000}\times 0.01+\dfrac{4000}{12000}\times 0.03+\dfrac{6000}{12000}\times 0.15 \\
 & \Rightarrow P\left( A \right)=\dfrac{1}{6}\times 0.01+\dfrac{1}{3}\times 0.03+\dfrac{1}{2}\times 0.15 \\
 & \Rightarrow P\left( A \right)=0.0016+0.02+0.075 \\
 & \Rightarrow P\left( A \right)0.0866 \\
\end{align}$
Putting the values of $P\left( \dfrac{A}{S} \right)$ , P(S) and P(A) in equation (3), we get:
$\begin{align}
  & P\left( \dfrac{S}{A} \right)=\dfrac{0.01\times \dfrac{1}{6}}{0.0866}=\dfrac{0.01}{0.5196}=\dfrac{1}{51.96}\approx \dfrac{1}{52} \\
 & \\
\end{align}$
So, the probability that a driver meeting with an accident is a scooter driver is $.$ .
Hence, option (a) is the correct answer.

Note: Students should note here that $P\left( A\cap S \right)$ is equal to $P\left( S\cap A \right)$ . So, we can substitute the value of $P\left( S\cap A \right)$ as P $P\left( \dfrac{A}{S} \right)$ .P(S) in equation (2) and proceed.