A cadiadate has to reach the examination centre in time, Probality of him going by bus or scooter or by other means of transport are `(3)/(10),(1)/(10) ,(3)/(5)` respectively. The probability of getting late , if the travels by bus is 1/4 ,1/3 if he travells by scooter and 0 for any other medium. But he reaches in time if the uses any mode of transport . He reached late at the centre. The probability that he travelled by bus is -

Options

A. `(1)/(9)`
B. `(2)/(13)`
C. `(9)/(13)`
D. None of these

Explanation

To solve the problem step by step, we will use the concept of conditional probability and the law of total probability.

### Step 1: Define Events
Let:
- \( A \): Event that the candidate travels by bus.
- \( B \): Event that the candidate travels by scooter.
- \( C \): Event that the candidate travels by other means.
- \( E \): Event that the candidate is late.

### Step 2: Given Probabilities
From the problem, we have the following probabilities:
- \( P(A) = \frac{3}{10} \) (Probability of traveling by bus)
- \( P(B) = \frac{1}{10} \) (Probability of traveling by scooter)
- \( P(C) = \frac{3}{5} \) (Probability of traveling by other means)

### Step 3: Calculate Total Probability
Since the total probability must equal 1, we can check:
\[ P(A) + P(B) + P(C) = \frac{3}{10} + \frac{1}{10} + \frac{3}{5} = \frac{3}{10} + \frac{1}{10} + \frac{6}{10} = 1 \]
This confirms the probabilities are correct.

### Step 4: Given Conditional Probabilities of Being Late
We also have the conditional probabilities of being late:
- \( P(E|A) = \frac{1}{4} \) (Probability of being late given he travels by bus)
- \( P(E|B) = \frac{1}{3} \) (Probability of being late given he travels by scooter)
- \( P(E|C) = 0 \) (Probability of being late given he travels by other means)

### Step 5: Use the Law of Total Probability
We need to find the total probability of being late \( P(E) \):
\[ P(E) = P(A) \cdot P(E|A) + P(B) \cdot P(E|B) + P(C) \cdot P(E|C) \]
Substituting the values:
\[ P(E) = \left(\frac{3}{10} \cdot \frac{1}{4}\right) + \left(\frac{1}{10} \cdot \frac{1}{3}\right) + \left(\frac{3}{5} \cdot 0\right) \]
Calculating each term:
\[ P(E) = \frac{3}{40} + \frac{1}{30} + 0 \]

### Step 6: Find a Common Denominator
To add \( \frac{3}{40} \) and \( \frac{1}{30} \), we find the least common multiple of 40 and 30, which is 120.
\[ \frac{3}{40} = \frac{3 \times 3}{40 \times 3} = \frac{9}{120} \]
\[ \frac{1}{30} = \frac{1 \times 4}{30 \times 4} = \frac{4}{120} \]
Now, adding these:
\[ P(E) = \frac{9}{120} + \frac{4}{120} = \frac{13}{120} \]

### Step 7: Find the Probability of Traveling by Bus Given Late
We want to find \( P(A|E) \) using Bayes' theorem:
\[ P(A|E) = \frac{P(E|A) \cdot P(A)}{P(E)} \]
Substituting the values:
\[ P(A|E) = \frac{P(E|A) \cdot P(A)}{P(E)} = \frac{\left(\frac{1}{4}\right) \cdot \left(\frac{3}{10}\right)}{\frac{13}{120}} \]
Calculating the numerator:
\[ \frac{1}{4} \cdot \frac{3}{10} = \frac{3}{40} \]
Now substituting:
\[ P(A|E) = \frac{\frac{3}{40}}{\frac{13}{120}} = \frac{3}{40} \cdot \frac{120}{13} = \frac{3 \cdot 3}{13} = \frac{9}{13} \]

### Final Answer
Thus, the probability that he traveled by bus given that he reached late is:
\[ \boxed{\frac{9}{13}} \]