A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus , scooter or by other means of transportation are respectiely `3/10,1/5,1/10` or `2/5`. The probabilities that he will be late are `1/4,1/3` and `1/12` if he comes by train, bus and scooter respectively but if he comes by other means of transport, then he will not be late. If the arrives late, find the probability that he comes by train.

Explanation

To solve the problem, we need to find the probability that the doctor comes by train given that he is late. We will use Bayes' theorem for this purpose.

### Step 1: Define the events
Let:
- \( T \): The event that the doctor comes by train.
- \( B \): The event that the doctor comes by bus.
- \( S \): The event that the doctor comes by scooter.
- \( O \): The event that the doctor comes by other means.
- \( L \): The event that the doctor is late.

### Step 2: Write down the given probabilities
From the problem, we have:
- \( P(T) = \frac{3}{10} \)
- \( P(B) = \frac{1}{5} = \frac{2}{10} \)
- \( P(S) = \frac{1}{10} \)
- \( P(O) = \frac{2}{5} = \frac{4}{10} \)

The total probability must sum to 1:
\[ P(T) + P(B) + P(S) + P(O) = 1 \]
\[ \frac{3}{10} + \frac{2}{10} + \frac{1}{10} + \frac{4}{10} = 1 \]

### Step 3: Write down the conditional probabilities of being late
- \( P(L|T) = \frac{1}{4} \)
- \( P(L|B) = \frac{1}{3} \)
- \( P(L|S) = \frac{1}{12} \)
- \( P(L|O) = 0 \) (he will not be late if he comes by other means)

### Step 4: Calculate the total probability of being late, \( P(L) \)
Using the law of total probability:
\[ P(L) = P(L|T)P(T) + P(L|B)P(B) + P(L|S)P(S) + P(L|O)P(O) \]
Substituting the values:
\[ P(L) = \left(\frac{1}{4} \cdot \frac{3}{10}\right) + \left(\frac{1}{3} \cdot \frac{2}{10}\right) + \left(\frac{1}{12} \cdot \frac{1}{10}\right) + \left(0 \cdot \frac{4}{10}\right) \]
Calculating each term:
\[ = \frac{3}{40} + \frac{2}{30} + \frac{1}{120} + 0 \]

Finding a common denominator (which is 120):
\[ = \frac{3 \cdot 3}{120} + \frac{2 \cdot 4}{120} + \frac{1}{120} \]
\[ = \frac{9}{120} + \frac{8}{120} + \frac{1}{120} = \frac{18}{120} = \frac{3}{20} \]

### Step 5: Use Bayes' theorem to find \( P(T|L) \)
According to Bayes' theorem:
\[ P(T|L) = \frac{P(L|T)P(T)}{P(L)} \]
Substituting the values:
\[ P(T|L) = \frac{\left(\frac{1}{4}\right) \left(\frac{3}{10}\right)}{\frac{3}{20}} \]
Calculating the numerator:
\[ = \frac{3}{40} \]
Now, substituting into Bayes' theorem:
\[ P(T|L) = \frac{\frac{3}{40}}{\frac{3}{20}} = \frac{3}{40} \cdot \frac{20}{3} = \frac{20}{40} = \frac{1}{2} \]

### Final Answer
The probability that the doctor comes by train given that he is late is \( \frac{1}{2} \).