QUESTION

An insurance company insured 2000 scooters and 3000 motorcycles. The probability of an accident involving a scooter is 0.01 and that of a motorcycle is 0.02. An insured vehicle met with an accident. Find the probability that the accidented vehicle was a motorcycle.

Hint: In this question, we will first find the probability of insurance of a scooter and motorcycle. Then, use it with the other given data and by applying Bayes theorem, we will find the required probability.

Let the event of insurance of a scooter be represented by E and the event of insurance of a motorcycle be represented by F.

Let A be the event that an accident has occurred.

Then, P(E) will be the probability of insurance of a scooter and P(F) will be the probability of insurance of a motorcycle.

Let P(A|E) be the probability of occurrence of A such that E has already occurred. And, P(A|F) be the probability of occurrence of A such that F has already occurred.

Now, to find P(F|A), that is, the probability of occurrence of F such that A has already occurred, we will use Bayes theorem.

According to Bayes theorem,

If E and F are two mutually disjoint events with P(E), P(F) ≠ 0, then for any arbitrary event A which is a subset of the union of events E and F, such that P(A) > 0, we have,
$P(A|F)=\dfrac{P\left( A|F \right).P(F)}{P(A)}$.

Now, from the question, the number of scooters insured is 2000 and the number of motorcycle insured is 3000. Therefore, total number of insurances = 2000+3000 = 5000.

Therefore, $P\left( E \right)=\dfrac{2000}{5000}=\dfrac{2}{5}=0.4$.

And, $P\left( F \right)=\dfrac{3000}{5000}=\dfrac{3}{5}=0.6$.

Also,

Probability of an accident involving a scooter = P(A|E) = 0.01.

And, probability of an accident involving a motorcycle = P(A|F) = 0.01.

Also, P(A) is given by the formula,

$P\left( A \right)=P\left( E \right)P\left( A|E \right)+P\left( F \right)P\left( A|F \right)\cdots \cdots \left( i \right)$.

Therefore, putting values of P(E), P(F), P(A|E) and P(A|F), we get,

\begin{align} & P\left( A \right)=0.4\times 0.01+0.6\times 0.02 \\ & =0.004+0.012 \\ & =0.016 \\ \end{align}

Hence, using Bayes theorem, the probability that the accidented vehicle was a motorcycle,
that is P(F|A), is given by,

$P(A|F)=\dfrac{P\left( A|F \right).P(F)}{P(A)}$

Putting the values calculated above,

\begin{align} & P(A|F)=\dfrac{0.02\times 0.6}{0.016} \\ & =\dfrac{0.012}{0.016} \\ & =\dfrac{12}{16} \\ & =\dfrac{3}{4} \\ \end{align}

Hence, the probability that the accidented vehicle was a motorcycle is $\dfrac{3}{4}$.

Note: In such a type of question, when events are mutually independent and the probability of one event is to be calculated depending on occurrence or non-occurrence of some other event, then we use Bayes theorem.