Q.An insurance company insured $2000$ scooter drivers, $4000$ car drivers and $6000$ truck drivers. The probability of an accidents are $0.01$, $0.03$ and $0.15$ respectively. One of the insured persons meets with an accident. What is the probability that he is a scooter driver?

Concept: Conditional Probability (Bayes' Theorem)

We are given the total number of insured drivers:

• Scooter: $2000$
• Car: $4000$
• Truck: $6000$

Total insured persons = $2000 + 4000 + 6000 = 12000$.

Let $S$, $C$, $T$ be the events that the person is a scooter, car, or truck driver respectively. Let $A$ be the event of an accident.

Step 1: Prior probabilities

Step 2: Likelihoods (accident probabilities given driver type)

Step 3: Apply Bayes' Theorem

Substitute:

Compute numerator: $\frac{1}{6} \times 0.01 = \frac{0.01}{6} = \frac{1}{600}$.

Denominator: …

Why Bayes' Theorem?

We are told that an accident has happened, and we need the probability that the person involved is a scooter driver. This is a classic case of inverse probability: we know the probabilities of accidents given each type of driver, but we want the probability of the driver type given that an accident occurred.

The natural tool is Bayes' Theorem. It lets us "reverse" the conditional probability using the prior probabilities (how many of each driver type are insured) and the likelihoods (accident rates for each type).

The denominator $P(\text{Accident})$ is the total probability of an accident across all driver types — we compute it using the law of total probability.

Step-by-step solution

1. Find the prior probabilities (proportions of insured drivers)

Total insured persons:

So:

• $P(\text{Scooter}) = \frac{2000}{12000} = \frac{1}{6}$
• $P(\text{Car}) = \frac{4000}{12000} = \frac{1}{3}$
• $P(\text{Truck}) = \frac{6000}{12000} = \frac{1}{2}$

2. Note the likelihoods (accident probabilities given driver type)

These are given directly:

• $P(\text{Accident} \mid \text{Scooter}) = 0.01$
• $P(\text{Accident} \mid \text{Car}) = 0.03$
• $P(\text{Accident} \mid \text{Truck}) = 0.15$

3. Compute the total probability of an accident

Using the law of total probability:

Let's compute each term carefully:

• Scooter: $0.01 \times \frac{1}{6} = \frac{0.01}{6} = \frac{1}{600}$
• Car: $0.03 \times \frac{1}{3} = 0.01 = \frac{1}{100}$
• Truck: $0.15 \times \frac{1}{2} = 0.075 = \frac{3}{40}$

Now add them. Convert to a common denominator of 600:

• $\frac{1}{600}$ stays as is
• $\frac{1}{100} = \frac{6}{600}$
• $\frac{3}{40} = \frac{45}{600}$

So: …

Method: Bayes' Theorem when the priors come as counts

Use this "given an event, find which category" pattern when the categories are described by group sizes (counts) rather than ready-made probabilities.

Steps

Step 1: Convert the counts into prior probabilities.

Add all group sizes to get the total, then $P(H_i) = \dfrac{\text{size of group } i}{\text{total}}$. Never plug the raw counts straight into the formula — they must become fractions that sum to $1$.

Step 2: Record each group's likelihood of the event.

$P(E \mid H_i)$ is the event rate (accident probability) for that group, given directly.

Step 3: Total probability of the event. …

Common Mistakes

Mistake 1: Using the raw counts (2000, 4000, 6000) as probabilities.

Why it's wrong: probabilities must be fractions of the total that sum to $1$. Correct approach: divide by $12000$ to get priors $\tfrac16,\tfrac13,\tfrac12$.

Mistake 2: Comparing accident rates without weighting by group size.

Why it's wrong: trucks are both the largest group and the highest-risk, so they dominate the accidents. Correct approach: weight each rate by its prior inside $P(A)$.

Mistake 3: Forgetting the full denominator $P(A)$. …