 Hello and welcome to the session. I am Lipika here. Let's discuss the question it says. An insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident are 0.01, 0.03 and 0.15 respectively. One of the insured persons meet with an accident. What is the probability that he is a scooter driver? Now, here we will use Bayes theorem according to this theorem. If e1, e2, so on, en are events which constitute a partition of central space s. That is, e1, e2, so on, en are pair-wise disjoint. And even union e2, union so on, union en is equal to s. And a be any event with non-zero probability, then probability of ei upon a is equal to probability of ei into probability of a upon ei over sigma probability of ej into probability of a upon ej, j varying from 1 to n. So, this is a key idea behind that question. We will take the help of this key idea to solve the above question. So, let's start the solution. Now, according to the given question, an insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. So, let e1, e2, e3 be the events that the insured persons scooter driver or driver driver respectively. Therefore, probability of e1 is equal to 2000 over 12000 and this is equal to 1 over 6 because 2000 scooter drivers are insured and total insured persons are 12000. So, probability of e1 is equal to 2000 over 12000 which is equal to 1 over 6. Similarly, probability of e2 is equal to 4000 over 12000 which is equal to 1 over 3 and probability of e3 equal to 6000 over 12000 which is equal to 1 over 2. Again, we are given an insurance company insured 2000 scooter drivers, 4000 car drivers and 6000 truck drivers. The probability of an accident are 0.01, 0.03 and 0.15 respectively and one of the insured persons meet with an accident. We have to find the probability that he is a scooter driver. So, let a be the event that the insured person met with an accident. Probability of a upon e1 that is probability that the person met with an accident is a scooter driver. This is equal to 0.01 because it is given in the question. Again, probability of a upon e2 that is the probability that the person met with an accident is a car driver and this is equal to 0.03. Again, probability of a upon e3 that is probability that the person met with an accident is a truck driver. This is equal to 0.15. Now, the probability that the scooter driver met with an accident, this is given by probability of e1 upon a. So, by using Bayes theorem, we have probability of e1 upon a is equal to probability of e1 into probability of a upon e1 over probability of e1 into probability of a upon e1 plus probability of e2 into probability of a upon e2 plus probability of e3 into probability of a upon e3. Now, we have probability of e1 is equal to 1 over 6. Probability of e2 is equal to 1 over 3 and probability of e3 is equal to 1 over 2 and probability of a upon e1 is equal to 0.01. Probability of a upon e2 is equal to 0.03 and probability of a upon e3 is equal to 0.15. So, probability of e1 upon a is equal to 1 over 6 into 0.01 over 1 over 6 into 0.01 plus 1 over 3 into 0.03 plus 1 over 2 into 0.05. This is again equal to 1 over 6 into 0.01 over 1 over 6 into 1 into 0.01 plus 2 into 0.03 plus 3 into 0.15. This is again equal to 0.01 over 0.01 into 1 plus 6 plus 45 and this is equal to 1 over 52. Hence, the answer for the above question is 1 over 52. That is, the probability that the scooter driver met with an accident is 1 over 52. So, this completes our session. I hope the solution is clear to you. Bye and have a nice day.