 Hello and welcome to the session. In this session we are going to discuss the following question which says that two unbiased dice are thrown and the sum of the numbers which come upon the dice is observed. 1. Find the probability that the sum is even. 2. What are the odds in favor of getting sum a multiple of 3? 3. What are the odds against getting sum greater than 11? Let a be any event. Probability of occurrence of event a is given by number of favorable outcomes upon total number of outcomes denoted by m of a by m of s in favor of a is given by number of times a equals by number of times a does not occur. What's against a is given by number of times a does not occur by number of times a equals. This key idea let us proceed with the solution. As two dice are thrown there are 36 elements in the sample space. Sample space s is given by x y such that x y can take any value from 1 to 6. Therefore total number of elements in the sample space is 36. We shall find the probability that the sum is even. Let a be an event getting sum as an even number. Therefore a is given by the set of ordered pair of elements that is 1 1, 1 3, 1 5, 2 2, 2 4, 3 1, 3 3, 4 2, 4 4, 8 1, 8 3. Therefore number of elements of a is equal to 18 as there are 18 elements in the set a. Therefore probability of occurrence of event a that is of getting sum as an even number is given by number of favorable outcomes divided by total number of outcomes that is n of a by n of s that is n of a is 18, n of s is 36 which is given by 1 by 2. Therefore probability of getting an even sum is 1 by 2. Now we shall find odds in favor of getting sum a multiple of 3. Let b be an event of getting sum a multiple of 3. Therefore elements of b will be set of ordered pair of elements having sum a multiple of 3. That is 1 2, 2 1, 1 5, 5 1, 3 3, 2 4, 4 2, 8 3, 4 5, 5 4 and 6. Number of elements of b is given by 12. We can say the number of times b occurs is given by 12 then number of times b does not occur is given by total number of outcomes minus number of times b occurs. Total number of outcomes is given by 36 and number of times b occurs is given by 12. Which is equal to 24. Therefore number of times b does not occur is given by 24. We know that odds in favor of an event say a is given by number of times a occurs by number of times a does not occur. Therefore odds in favor of event b is given by number of times b occurs upon number of times b does not occur. Number of times b occurs is given by 12 and number of times b does not occur is given by 24. So we have 12 by 24 which is equal to 1 by 2. Therefore odds in favor of getting the sum as multiple of 3 is given by 1 by 2. Now we shall find odds against getting sum greater than 11. Let's see the event of getting sum greater than 11. Therefore c is given by the other pair of elements 6 of the sum of the elements is 12 which is greater than 11. Therefore number of favorable outcomes for the event c is 1. Or we can say number of times b occurs is 1. Then number of times c does not occur is given by total number of outcomes minus number of times c occurs which is given by 36 minus 1 that is 35. Therefore number of times c does not occur is given by 35. Now we know that odds against any event say a is given by number of times a does not occur by number of times a occurs. Therefore odds against even c is given by number of times c does not occur by number of times c occurs. Number of times c occurs is given by 1 and number of times c does not occur is given by 35. So we have 35 by 1 that is 35. Therefore odds against getting sum greater than 11 is given by 35. This completes our session. Hope you enjoyed this session.