 Hello and welcome to the session. In this session we will discuss a question which says that two rows are thrown. The events A, B and C are as follows. Event A is given an even number and first time event B is given the sum of numbers on the rows less than or equal to 5 and event C is given an odd number on exactly one of the rows. Now we have to describe the following events and also find their probabilities. First is A complement, second part A or B, third part B and C and fourth part is A union C all complement. Now let us start with the solution of the given question. First of all let us find the central space of the given experiment. Here we have thrown two dies. Now we know that in an experiment of rolling a die total number of outcomes that is getting a number 1, a number 2, number 3, number 4, number 5 and number 6. Now when two dies are rolled simultaneously when total number of outcomes will be getting a number 1 on first die and getting a number 1 on the second die then getting a number 1 on the first die and getting a number 2 on the second die then the next outcome will be getting a number 1 on the first die and getting a number 3 on the second die and continuing this way we get these outcomes when we roll two dies simultaneously. Thus central space S is the set containing all these ordered pairs. Now the total number of ordered pairs is equal to 36. This means number of elements in central space is equal to 36. Now it is given that event A is getting an even number on first die. This means only 2, 4 and 6 numbers can appear on first die. This means number of related outcomes for event A will be the ordered pairs with first components 2, 4 or 6. So A is the set containing the ordered pairs 2, 1, 2, 2, 2, 3, 2, 4, 2, 5, 2, 6, 4, 1, 4, 2, 4, 3, 4, 4, 4, 4, 5, 4, 6, 6, 1, 6, 2, 6, 3, 6, 4, 6, 5, 6, 6. So when we count these ordered pairs in set A then we get the number of favorable outcomes for event A which is equal to 18. Now event B is getting the sum of numbers on the dies less than equal to 5. Now here we can see the least sum of numbers on the dies is 1 plus 1 that is 2. Now event B is getting sum of numbers on the dies less than or equal to 5. Thus a total can be 2, 3, 4 or 5. So B is a set containing the ordered pairs 1, 1, 1, 2, 1, 3, 1, 4, 2, 1, 2, 2, 2, 3, 3, 1, 3, 2 and 4, 1. Now here you can see 1 plus 1 is 2, 1 plus 2 is 3, 1 plus 3 is 4, 1 plus 4 is 5, 2 plus 1 is 3, 2 plus 2 is 4, 2 plus 3 is 5, 3 plus 1 is 4, 3 plus 2 is 5 and 4 plus 1 is so number of favorable outcomes for event B is equal to 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10. Now event C is getting an odd number on exactly 1 of the dies. Now here we must have only one odd number on one of the two dies. So C is a set containing the ordered pairs 1, 2, 1, 4, 1, 6, 2, 1, 2, 3, 2, 5, 3, 2, 3, 4, 3, 6, 4, 5, 6, 4, 1, 4, 3, 4, 5, 5, 2, 5, 4, 6, 1, 6, 3, 6, 5. So here number of favorable outcomes for event C is equal to 18. Now we have to find A complement. Now A complement will have those elements of surface space which are not in A. All you can say set A complement will have all those elements of set S which are not in set A. Now leaving all the elements listed in set A, the remaining elements will be in set A complement. So A complement is a set containing the ordered pairs 1, 1, 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 3, 1, 3, 2, 3, 3, 3, 4, 3, 5, 3, 6, 5, 1, 5, 2, 5, 3, 5, 4, 5, 5 and 5, 6. Now number of set A complement is equal to 18. It means number of favorable outcomes for event A complement is equal to 18. Now we have to find probability of event A complement. Now we know that probability of an event E at this P of E is equal to number of favorable outcomes for event E upon total number of outcomes. So probability of event A complement will be equal to number of favorable outcomes for event A complement that is 18 upon total number of outcomes that is number of elements in sample space which is 36. So probability of event A complement is equal to 18 upon 36 which is equal to 1 upon 2. Now in the second part we have to find A or B. Now A or B means union of two events. So we take all elements in event A and in event B and repeated elements are written only once thus A or B is equal to set containing order pairs 1, 1, 1, 2, 1, 3, 1, 4, 2, 1, 2, 2, 2, 3, 2, 4, 2, 5, 2, 6, 3, 1, 3, 2, 4, 1, 4, 2, 4, 3, 4, 4, 4, 5, 4, 6, 6, 1, 6, 2, 6, 3, 6, 4 is equal to 24. Now let us find probability P of an event A union B which is equal to number of elements favorable to event A union B which is 24 upon total number which is 36. So this is equal to 2 upon 3. Now in the third part we have to find event B and C. Now event B and C means intersection of events that is very wide to both B and C and C is a set containing order pairs 1, 2, 1, 4, 2, 1, 2, 3, 3, 2, and 4, 1. So number of elements in the set B intersection C is equal to 1, 2, 3, 4, 5 and 6. It means number of variable outcomes for event B and C is equal to so probability P of event B intersection C is equal to number of elements favorable to event B intersection C that is 6 upon total number of elements that is 36 and this is equal to 1 upon 6. Now in the next part we have to find event A union C whole complement. It will include those elements which are neither in set A nor in set C. So from surface space we write the elements leaving the elements of set A and set C thus in union C whole complement is equal to set containing the order pairs 1, 1, 1, 3, 1, 5, 3, 1, 3, 3, 5, 1, 5, 5. Number of other union C whole complement is equal to 1, 2, 3, 4, 5, 6, 7, 8 and 9. Thus probability P of event A union C whole complement is equal to 9 upon 36 which is equal to 1 upon 4. So this is the solution of the given question and that's all for this session. Hope you all have enjoyed the session.