Question

The probability that a person goes to college by car,\(\frac{1}{5}\) bus,\(\frac{2}{5}\) or train \(\frac{3}{5}\) is given. If he reaches college on time, find the probability he traveled by car.

• A. \( \frac{6}{29} \)
• B. \( \frac{24}{29} \)
• C. \( \frac{5}{29} \)
• D. \( \frac{23}{29} \)

Hint:

Use Bayes' theorem for conditional probability problems involving different cases.

Answer 1

The Correct Option is C

Step 1: Given Data
The probability of taking a particular mode of transport: \[ P(C) = \frac{1}{5}, \quad P(B) = \frac{2}{5}, \quad P(T) = \frac{3}{5} \] The probability of reaching late for each mode: \[ P(L | C) = \frac{2}{7}, \quad P(L | B) = \frac{4}{7}, \quad P(L | T) = \frac{1}{7} \] Step 2: Finding Probability of Reaching on Time
\[ P(T) = 1 - P(L) \] \[ P(L) = P(C) P(L | C) + P(B) P(L | B) + P(T) P(L | T) \] \[ = \left(\frac{1}{5} \times \frac{2}{7} \right) + \left(\frac{2}{5} \times \frac{4}{7} \right) + \left(\frac{3}{5} \times \frac{1}{7} \right) \] \[ = \frac{2}{35} + \frac{8}{35} + \frac{3}{35} = \frac{13}{35} \] \[ P(\text{on time}) = 1 - \frac{13}{35} = \frac{22}{35} \] Step 3: Using Bayes' Theorem
\[ P(C | T) = \frac{P(C) P(T | C)}{P(T)} \] \[ = \frac{\left(\frac{1}{5} \times \frac{5}{7}\right)}{\frac{22}{35}} \] \[ = \frac{5}{35} \times \frac{35}{29} = \frac{5}{29} \] Thus, the correct answer is \( \frac{5}{29} \).