 Hello and welcome to this session. In this session we will discuss a question which says that a sample contains 20 belts. In the sample, 4 belts are defective. If 3 belts are selected as random, there is the probability that exactly 2 belts are defective. And what is the probability that all belts are defective? Now let us start with the solution of the given question. Now here we have given a sample of 20 belts and out of these 3 belts are selected. So the total number of outcomes will be chosen 3 out of 20 belts that is 20 C3 which means number of combinations of 20 belts taken 3 at a time. So total number of outcomes is equal to 20 C3. Now we know that NCR is equal to N factorial upon R factorial into N minus R whole factorial. So 20 C3 will be equal to 20 factorial upon 3 factorial into 20 minus 3 whole factorial which is equal to 20 factorial upon 3 factorial into 20 minus 3 is 17. So this will be 17 factorial. Now 20 factorial can be written as 20 into 19 into 18 into 17 factorial whole whole. Now 3 factorial is 3 into 2 into 1 into 17 factorial further in solving. This is equal to 20 into 19 into 3 which is equal to 1140. Therefore total number of outcomes is equal to 1140. Now in the first part we have to find probability that exactly 2 belts are defective. Now here we know that these are selected and out of these 3 belts we have to find the probability that exactly 2 belts are defective. It means we choose 1 non defective belts. Now out of 20 belts number of defective belts are 4 and non defective belts will be 20 minus 4 that is equal to 16 belts. So here we have to choose 2 defective belts out of 7 defective belts, 1 non defective belts out of still non defective belts. Thus number of variable outcomes for this event that is 2 defective and 1 non defective belts will be choosing 2 out of 4 defective belts that is 4C2 into choosing 1 out of 16 non defective belts that is 16C1. So this is equal to now 4C2 will be 4 factorial upon 2 factorial into 4 minus 2 factorial into 16C1 will be 16 factorial upon 1 factorial into 16 minus 1 factorial. Further this is equal to 4 factorial upon 2 factorial into 2 factorial into 16 factorial upon 1 factorial is 1 and this will be 15 factorial. Now this is equal to now 4 factorial can be written as 4 into 3 into 2 factorial upon 2 factorial can be written as 2 into 1 into 2 factorial into 16 factorial can be written as 16 into 15 factorial upon 15 factorial. Now further on solving this is equal to 6 into 16 which is equal to 96. Therefore number of factorial outcomes is equal to 96. Now we have to find probability P that exactly 2 defective belts are selected. Now we know that probability P of an event E is equal to number of favorable outcomes for event E upon total number of outcomes. Now here number of favorable outcomes for this event is 96 and total number of outcomes is equal to 1130. So probability that exactly 2 defective belts are selected is equal to 96 upon 1140. Now we know that 4 into 24 is 96 and 4 into 285 is 1140. So this is equal to 24 upon 285. Now in the second part we have to find the probability that all belts are defective. Now we know that 3 belts are selected as random and here we have to find the probability that all belts are defective. And we also know that out of 20 belts number of defective belts is 4 and number of non-defective belts is 16. So favorable outcomes for this event are choosing 3 defective belts out of 4 defective belts and choosing 0 non-defective belts out of 16 non-defective belts. So number of favorable outcomes for this event is equal to 4C3 into 16C0. Now this is equal to 4 factorial upon 3 factorial into 4 minus 3 whole factorial into 16 factorial upon 0 factorial into 16 minus 0 whole factorial. Now we know that 0 factorial is equal to 1. So this is equal to 4 factorial upon 3 factorial into now 4 minus 3 is 1 factorial into 16 factorial upon 0 factorial is 1 into 16 minus 0 whole factorial. That is 16 factorial further this is equal to now 4 factorial can be written as 4 into 3 factorial and we know that 1 factorial is 1. So 3 factorial into 1 is 3 factorial into now here on calculating this is equal to 1 further this is equal to 4 into 1 that is 4. So number of favorable outcomes for this event is equal to 4. Now we will have to find probability P that all defective belts are selected. Now number of favorable outcomes for this event is equal to 4 comes is equal to 1140. So probability that all defective belts are selected is equal to 4 upon 1140. Now 4 into 285 is 1140. So this is equal to 1 upon 285. So this is the solution of the given question and that's all for this session. Hope you all have enjoyed the session.