Q.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 transport are respectively $\frac{3}{10}$, $\frac{1}{5}$, $\frac{1}{10}$ and $\frac{2}{5}$. The probabilities that he will be late are $\frac{1}{4}$, $\frac{1}{3}$ and $\frac{1}{12}$, if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train?

Concept: Conditional Probability (Bayes' Theorem)

We are given the prior probabilities of each mode of transport and the conditional probabilities of being late given each mode. The doctor is late — we need the posterior probability that he came by train.

Step 1: Define events

Let $T$, $B$, $S$, $O$ be the events that he comes by train, bus, scooter, or other means.

$P(T) = \frac{3}{10},\; P(B) = \frac{1}{5},\; P(S) = \frac{1}{10},\; P(O) = \frac{2}{5}$.

Let $L$ = event that he is late.

$P(L|T) = \frac{1}{4},\; P(L|B) = \frac{1}{3},\; P(L|S) = \frac{1}{12},\; P(L|O) = 0$.

Step 2: Compute total probability of being late

$= \frac{3}{10}\cdot\frac{1}{4} + \frac{1}{5}\cdot\frac{1}{3} + \frac{1}{10}\cdot\frac{1}{12} + \frac{2}{5}\cdot 0$ …

The problem gives us a classic Bayes’ theorem situation: we know the prior probabilities of which transport the doctor uses, and we know the conditional probabilities of being late given each transport. When we observe that he is late, we want the posterior probability that he came by train.

The key idea: Bayes’ theorem reverses the conditioning. Instead of asking “If he takes the train, how likely is he to be late?” we ask “Given that he is late, how likely is it that he took the train?” The formula is:

The denominator $P(\text{Late})$ is the total probability of being late, summed over all possible transport modes.

Let’s work through it step by step.

1. List the given probabilities clearly.

   Let $T$, $B$, $S$, and $O$ denote the events that the doctor comes by train, bus, scooter, or other means, respectively.

   • $P(T) = \frac{3}{10}$
   • $P(B) = \frac{1}{5} = \frac{2}{10}$
   • $P(S) = \frac{1}{10}$
   • $P(O) = \frac{2}{5} = \frac{4}{10}$

   Check: $\frac{3}{10} + \frac{2}{10} + \frac{1}{10} + \frac{4}{10} = 1$, good.

2. Conditional probabilities of being late.

   Let $L$ be the event that the doctor is late.

   • $P(L \mid T) = \frac{1}{4}$
   • $P(L \mid B) = \frac{1}{3}$
   • $P(L \mid S) = \frac{1}{12}$
   • $P(L \mid O) = 0$ (he will not be late by other means)

3. Compute the total probability of being late, $P(L)$.

   By the law of total probability:

Substitute:

Compute each term:

• $\frac{1}{4} \cdot \frac{3}{10} = \frac{3}{40}$
• $\frac{1}{3} \cdot \frac{1}{5} = \frac{1}{15}$
• $\frac{1}{12} \cdot \frac{1}{10} = \frac{1}{120}$

Now add them. Find a common denominator — the LCM of 40, 15, and 120 is 120:

• $\frac{3}{40} = \frac{9}{120}$
• $\frac{1}{15} = \frac{8}{120}$
• $\frac{1}{120} = \frac{1}{120}$

So:

Method: Bayes' Theorem (finding the cause from the observed outcome)

Use this when an outcome (being late) can arise through several channels and you want the probability of a particular channel given the outcome. You know the chance of being late given each mode of transport, but want the transport given that lateness occurred.

Steps

Step 1: Set up the partition of possible causes.

List the mutually exclusive, exhaustive hypotheses (here: he came by train, bus, scooter or other means) and write their prior probabilities $P(E_i)$ (these must sum to $1$).

Step 2: Write each likelihood.

For each cause, state $P(A\mid E_i)$ — the probability of the observed outcome (here: the doctor is late) under that cause.

Step 3: Get the total probability of the outcome (the denominator).

By the law of total probability,

Step 4: Apply Bayes' theorem for the cause you want.

$P(E_k\mid A)=\frac{P(E_k)\,P(A\mid E_k)}{\sum_i P(E_i)\,P(A\mid E_i)}.$ …

Common Mistakes

Mistake 1: Dropping the "other means" mode from the partition.

Why it's wrong: even though $P(\text{late}\mid\text{other})=0$, the mode's prior $P(O)=\tfrac25$ is needed for the priors to sum to $1$, and forgetting the mode entirely signals a mis-set partition. Correct approach: list all four modes; its late-contribution is simply zero.

Mistake 2: Using $P(\text{late}\mid\text{train})=\tfrac14$ directly as the answer.

Why it's wrong: that is the likelihood, not $P(\text{train}\mid\text{late})$. Correct approach: apply Bayes' theorem with the full total-probability denominator. …