Question

A person goes to office by a car or scooter or bus or train, probability of which are $\dfrac{1}{7}$, $\dfrac{3}{7}$,$\dfrac{2}{7}$and $\dfrac{1}{7}$respectively. Probability that he reaches office late, if he takes car, scooter, bus or train is $\dfrac{2}{9},\dfrac{1}{9},\dfrac{4}{9},$and $\dfrac{1}{9}$ respectively. Given that he reached the office in time, the probability he travelled by a car is

& \text{(A) }\dfrac{1}{7} \\\ & (\text{B) }\dfrac{2}{7} \\\ & (\text{C) }\dfrac{3}{7} \\\ & (\text{D) }\dfrac{4}{7} \\\ \end{aligned}$$

Answer 1

We should have an idea about Bayes Theorem to solve this problem. From Bayes| Theorem,
We know that $P\left( \dfrac{{{E}_{i}}}{A} \right)=\dfrac{P({{E}_{i}})P\left( \dfrac{A}{{{E}_{i}}} \right)}{P(A)}$ where $P\left( \dfrac{{{E}_{i}}}{A} \right)$ is the probability of occurrence of event ${{E}_{i}}$after occurrence of event A, $P\left( \dfrac{A}{{{E}_{i}}} \right)$ is the probability of occurrence of event A after occurrence of event ${{E}_{i}}$, $P({{E}_{i}})$ is the probability of occurrence of event ${{E}_{i}}$ and $P(A)=P({{E}_{1}})P\left( \dfrac{A}{{{E}_{1}}} \right)+P({{E}_{2}})P\left( \dfrac{A}{{{E}_{2}}} \right)+........+P({{E}_{\lambda }})P\left( \dfrac{A}{{{E}_{\lambda }}} \right)$ where ${{E}_{1}},{{E}_{2}},{{E}_{3}},.....,{{E}_{n}}$ are n events. By using this concept, we can solve this problem.

Complete step by step answer:
Let us assume
Probability that the person goes to office by car is equal to P(C).
Probability that the person goes to the office by scooter is equal to P(S).
Probability that the person goes to office by bus is equal to P(B).
Probability that the person goes to office by train is equal to P(T).
Probability that he reaches the office late if he takes a car is equal to $P({{C}_{1}})$.
Probability that he reaches the office late if he takes a scooter is equal to $P({{S}_{1}})$ .
Probability that he reaches the office late if he takes bus is equal to $P({{B}_{1}})$.
Probability that he reaches the office late if he takes the train is equal to $P({{T}_{1}})$.
Probability that he reaches the office in time if he takes the car is equal to $P\left( {{C}_{1}}' \right)$.
Probability that he reaches the office in time if he takes a scooter is equal to $P\left( {{S}_{1}}' \right)$.
Probability that he reaches the office in time if he takes the bus is equal to $P\left( {{B}_{1}}' \right)$.
Probability that he reaches the office in time if he takes the train is equal to $P\left( {{T}_{1}}' \right)$.
In the question, it was given that
Probability that the person goes to office by car is equal to $\dfrac{1}{7}$.
Probability that the person goes to the office by scooter is equal to $\dfrac{3}{7}$.
Probability that the person goes to office by bus is equal to $\dfrac{2}{7}$.
Probability that the person goes to office by train is equal to $\dfrac{1}{7}$.
Probability that he reaches the office late if he takes a car is equal to $\dfrac{2}{9}$.
Probability that he reaches the office late if he takes a scooter is equal to $\dfrac{1}{9}$.
Probability that he reaches the office late if he takes the bus is equal to $\dfrac{4}{9}$.
Probability that he reaches the office late if he takes the train is equal to $\dfrac{1}{9}$.
So,
Probability that he reaches the office in time if he takes the car $=1-\dfrac{2}{9}=\dfrac{7}{9}$.
Probability that he reaches the office in time if he takes a scooter $=1-\dfrac{1}{9}=\dfrac{8}{9}$.
Probability that he reaches the office in time if he takes the bus $=1-\dfrac{4}{9}=\dfrac{5}{9}$.
Probability that he reaches the office in time if he takes the train $=1-\dfrac{1}{9}=\dfrac{8}{9}$.
From the question it was given that we have to find the probability that the person reaches the office in time by using a car.
So, it is clear that we have to find the value of $P\left( \dfrac{{{C}_{1}}'}{A} \right)$.
From conditional probability concept, we get that
Now by using the conditional probability concept, we can find the total probability that the person can reach the office by time.
$P(A)=P({{E}_{1}})P\left( \dfrac{A}{{{E}_{1}}} \right)+P({{E}_{2}})P\left( \dfrac{A}{{{E}_{2}}} \right)+........+P({{E}_{\lambda }})P\left( \dfrac{A}{{{E}_{\lambda }}} \right)$ where ${{E}_{1}},{{E}_{2}},{{E}_{3}},.....,{{E}_{n}}$ are n events.
From Bayes Theorem,
We know that $P\left( \dfrac{{{E}_{i}}}{A} \right)=\dfrac{P({{E}_{i}})P\left( \dfrac{A}{{{E}_{i}}} \right)}{P(A)}$ where $P\left( \dfrac{{{E}_{i}}}{A} \right)$ is the probability of occurrence of event ${{E}_{i}}$after occurrence of event A, $P\left( \dfrac{A}{{{E}_{i}}} \right)$ is the probability of occurrence of event A after occurrence of event ${{E}_{i}}$ and $P({{E}_{i}})$ is the probability of occurrence of event ${{E}_{i}}$.
By applying Bayes theorem, we can write the value of $P\left( \dfrac{{{C}_{1}}'}{A} \right)$.
So,
$P\left( \dfrac{{{C}_{1}}'}{A} \right)=\dfrac{P({{C}_{1}}')P\left( \dfrac{A}{{{C}_{1}}'} \right)}{P({{C}_{1}}')P\left( \dfrac{A}{{{C}_{1}}'} \right)+P({{T}_{1}})P\left( \dfrac{A}{T{{'}^{}}} \right)+P({{B}_{1}}')P\left( \dfrac{A}{{{B}_{1}}'} \right)+P({{S}_{1}}')P\left( \dfrac{A}{{{S}_{1}}'} \right)}$

& \Rightarrow P\left( \dfrac{{{C}_{1}}'}{A} \right)=\dfrac{\dfrac{1}{7}.\dfrac{7}{9}}{\dfrac{1}{7}.\dfrac{7}{9}+\dfrac{3}{7}.\dfrac{8}{9}+\dfrac{2}{7}.\dfrac{5}{9}+\dfrac{1}{7}.\dfrac{8}{9}} \\\ & \Rightarrow P\left( \dfrac{{{C}_{1}}'}{A} \right)=\dfrac{\dfrac{7}{63}}{\dfrac{7+24+10+8}{63}} \\\ & \Rightarrow P\left( \dfrac{{{C}_{1}}'}{A} \right)=\dfrac{\dfrac{7}{63}}{\dfrac{49}{63}}=\dfrac{7}{49} \\\ & \Rightarrow P\left( \dfrac{{{C}_{1}}'}{A} \right)=\dfrac{1}{7} \\\ \end{aligned}$$ So, the probability that he reached the office in time by travelling in a car is equal to $$\dfrac{1}{7}$$. **So, the correct answer is “Option A”.** **Note:** Students may make a mistake while calculating the probability. Generally, students may forget to find the probabilities of $$P\left( {{C}_{1}}' \right)$$, $$P\left( {{S}_{1}}' \right)$$, $$P\left( {{B}_{1}}' \right)$$ and $$P\left( {{T}_{1}}' \right)$$. They may use P (C), P(S), P(B) and P(T) instead of $$P\left( {{C}_{1}}' \right)$$, $$P\left( {{S}_{1}}' \right)$$, $$P\left( {{B}_{1}}' \right)$$ and $$P\left( {{T}_{1}}' \right)$$. This will give a wrong result. We must find the probability that he reaches the office in time. So, we use $$P\left( {{C}_{1}}' \right)$$, $$P\left( {{S}_{1}}' \right)$$, $$P\left( {{B}_{1}}' \right)$$ and $$P\left( {{T}_{1}}' \right)$$. To find the probability that he does not reach the office we have to use P (C), P(S), P(B) and P(T).