 Hi and welcome to the session. Let us discuss the following question. Question says, a lot of 20 bulbs contain 4 defective ones. One bulb is drawn at random from the lot. What is the probability that this bulb is defective? Second part is, suppose the bulb drawn in first part is not defective and is not replaced. Now, one bulb is drawn at random from the rest. What is the probability that this bulb is not defective? First upon, let us understand that probability of occurrence of an event E denoted by P e is defined as number of outcomes favourable to E upon total number of possible outcomes. This is the key idea to solve the given question. Let us now start with the solution. First upon, let us start with the first part of the given question. Now, we are given a lot of 20 bulbs and it contains 4 defective bulbs. We are given that one bulb is drawn at random from the lot of 20 bulbs. Now one bulb can be drawn in 20 ways. So, total number of possible outcomes is equal to 20. We are given number of defective bulbs is equal to 4. Now one defective bulb can be chosen in 4 ways. So, number of outcomes favourable to defective bulb is equal to 4. Now we have to find the probability of getting a defective bulb in one draw. It is equal to number of outcomes favourable to defective bulb that is 4 upon total number of possible outcomes that is 20. In this case, event is getting the defective bulb. So, probability of getting the defective bulb is equal to total number of outcomes favourable to defective bulb upon total number of possible outcomes that is 4 upon 20. Now we will cancel common factor 4 from numerator and denominator both and we get probability of getting a defective bulb in one draw is equal to 1 upon 5. So, required answer for the first part is probability that the bulb drawn is defective is equal to 1 upon 5. Let us now start with the second part. Now we are given that the bulb drawn in first part is not defective and it is not replaced. So, number of rest of the bulbs is equal to 20 minus 1 that is 19. We know we have picked up one bulb in first part of the given question and this one bulb drawn is not defective and it has not been replaced. So, rest of the bulbs equal to 20 minus 1 that is 19. So, we get number of rest of the bulbs is equal to 19. Now one bulb is drawn from rest of these bulbs. Now clearly we can see there are total 19 bulbs left and one bulb can be chosen in 19 ways. So, total number of possible outcomes is equal to 19. We know that out of these 19 bulbs 4 are defective. So, we can write number of defective bulbs is equal to 4 and number of non-defective bulbs is equal to 19 minus 4. We know total number of bulbs are 19 and number of defective bulbs are 4. So, if we subtract defective bulbs from the total bulbs we get number of non-defective bulbs. So, number of non-defective bulbs is equal to 15. Now one non-defective bulb can be chosen in 15 ways. So, number of outcomes favourable to non-defective bulb is equal to 15. Now we have to find the probability that the bulb drawn second time is not defective. So, probability of getting a non-defective bulb is equal to number of outcomes favourable to non-defective bulb that is 15 upon total number of possible outcomes that is 19. So, we get probability of getting non-defective bulb is equal to 15 upon 19. So, required answer for the second part of the given question is probability that bulb drawn is not defective is equal to 15 upon 19. So, this is our required answer. This completes the session. Hope you understood the solution. Take care and have a nice day.