Question

A person goes to college either by bus, scooter or car. The probability that he goes by bus is \(\frac{2}{5}\), by scooter is \(\frac{1}{5}\) and by car is \(\frac{3}{5}\). The probability that he entered late in college if he goes by bus is \(\frac{1}{7}\), by scooter is \(\frac{3}{7}\) and by car is \(\frac{1}{7}\). If it is given that he entered late in college, then the probability that he goes to college by car is

• A. \(\frac{3}{7}\)
• B. \(\frac{3}{8}\)
• C. \(\frac{4}{7}\)
• D. \(\frac{5}{8}\)

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This is a classic application of Bayes' Theorem, which allows us to update the probability of a cause (mode of transport) given that an event (being late) has occurred.
Bayes' Theorem states: \[ P(\text{Car} \mid \text{Late}) = \frac{P(\text{Car}) \cdot P(\text{Late} \mid \text{Car})}{P(\text{Late})} \] where the total probability of being late is: \[ P(\text{Late}) = P(B)\cdot P(L\mid B) + P(S)\cdot P(L\mid S) + P(C)\cdot P(L\mid C) \] Step 2: Listing All Given Data:
\[ P(B) = \frac{2}{5},\quad P(\text{Late}\mid B) = \frac{1}{7} \] \[ P(S) = \frac{1}{5},\quad P(\text{Late}\mid S) = \frac{3}{7} \] \[ P(C) = \frac{3}{5},\quad P(\text{Late}\mid C) = \frac{1}{7} \] Step 3: Computing $P(\text{Late})$ using Total Probability:
\[ P(\text{Late}) = \frac{2}{5}\cdot\frac{1}{7} + \frac{1}{5}\cdot\frac{3}{7} + \frac{3}{5}\cdot\frac{1}{7} \] \[ = \frac{2}{35} + \frac{3}{35} + \frac{3}{35} = \frac{8}{35} \] Step 4: Applying Bayes' Theorem:
\[ P(\text{Car}\mid\text{Late}) = \frac{P(C)\cdot P(\text{Late}\mid C)}{P(\text{Late})} = \frac{\dfrac{3}{5}\cdot\dfrac{1}{7}}{\dfrac{8}{35}} \] \[ = \frac{\dfrac{3}{35}}{\dfrac{8}{35}} = \frac{3}{8} \] Step 5: Final Answer:
The probability that the person went by car given he was late is $\dfrac{3}{8}$.
The answer is Option (2).