 Hello and welcome to the session. Let us discuss the following question which says, An insurance company insured 2,000 scooter drivers, 4,000 car drivers and 6,000 bike drivers. The probability of an accident involving a scooter, a car and a truck 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? To solve this question, we will use Bayes theorem according to which if E1, E2 and so on up to EN are n non-empty events which constitute a partition of sample space that is E1, E2 and so on up to EN are pairwise this joint E1 union E2 union and so on up to union EN is equal to s and a is any event of non-zero probability then probability of the event EI given the event A is equal to probability of the event EI into probability of the event A given the event EI upon sigma probability of the event EJ into probability of the event A given the event EJ when J varies from 1 to n for any I from 1, 2 and so on up to n. So this is the key idea for this question. Now let's see its solution. First of all we are given in the question that an insurance company insured 2,000 scooter drivers, 4,000 car drivers and 6,000 truck drivers. So let us define event E1 as a scooter driver is chosen event E2 as a car driver is chosen event E3 as a truck driver is chosen. Now let us find out the total number of persons insured this will be equal to 2,000 plus 4,000 plus 6,000 is equal to 12,000. Now the probability of the event E1 that is the probability that is scooter driver is chosen will be number of scooter drivers insured that is 2,000 upon total number of persons insured that is 12,000 and this will be equal to 1 upon 6. Similarly probability of the event E2 that is the probability that a car driver is chosen will be number of car drivers insured 4,000 upon total number of persons insured 12,000 and this will be equal to 1 upon 3. And probability of the event E3 that is the probability that a truck driver is chosen is number of truck drivers insured that is 6,000 upon total number of persons insured that is 12,000 is equal to 1 upon 2. Now we are given that the probability of an accident involving a scooter, a car and a truck are 0.01, 0.03 and 0.15 respectively. Now one of the insured persons meets with an accident and we need to find out the probability that the person who met with an accident is a scooter driver. Now let us assume that A with the event that a person meets with an accident. So we are given in the question the probability that a person who met an accident is a scooter driver which is given by probability of the event A given the event E1 is given to be 0.01. Similarly it is also given that the probability that a person met an accident is a car driver which is given by the probability of the event A given the event E2 is equal to 0.03 and the probability that the person met an accident is a truck driver given by the probability of the event A given the event E3 is equal to 0.15. Now we need to find out the probability that a scooter driver met with an accident which is given by the probability of the event E1 given the event A so by Bayes theorem we have probability of the event E1 given the event A is equal to the probability of the event E1 into the probability of the event A given the event E1 upon probability of the event E1 into probability of the event A given the event E1 plus probability of the event E2 into probability of the event A given the event E2 plus probability of the event E3 into probability of the event A given the event E3 So let us substitute all these values to find out the required probability. So this will be equal to 1 by 6 into 0.01 upon 1 by 6 into 0.01 plus 1 by 3 into 0.03 plus 1 by 2 into 0.15 which will be equal to 1 by 6 into 0.01 upon 1 by 6 into 0.01 plus 2 into 0.03 plus 3 into 0.15 that is 0.01 upon 0.01 plus 0.06 plus 0.45 which is equal to 0.01 upon 0.52 that is 1 upon 52. Thus 1 upon 52 is the required probability. With this we finish this session. Hope you must have understood the question. Goodbye, take care and have a nice day.