 Hi friends, I am Purva and today we will discuss the following question. A fair die is rolled. Consider events E which is equal to a set which consists of elements 1,3,5, F which is a set consisting of elements 2,3 and G which is a set consisting of elements 2,3,4,5. Find probability of E given F and find probability of F given E. Now if E and F are two events associated with the sample space of a random experiment, then the conditional probability of the event E given that F has occurred, that is probability of E given F is given by probability of E given F is equal to probability of E intersection F upon probability of F provided probability of F is not equal to 0. So this is the key idea behind our question. Let us begin with the solution now. Now let S be the sample space of the experiment of rolling a fair die. Then S is a set consisting of six elements 1,2,3,4,5,6. Now we are given that E is a set which consists of elements 1,3,5, F consists of 2,3 and G consists of 2,3,4,5. Now we have to find probability of E given F and probability of F given E. Now by key idea we know that probability of E given F is given by probability of E given F is equal to probability of E intersection F upon probability of F and similarly probability of F given E is equal to probability of F intersection E upon probability of E. We mark this as 1 and we mark this as 2. Now E is a set consisting of elements 1,3,5. So probability of E is equal to 3 upon 6 because total number of elements are 6 out of which 3 of them are in E. So we get probability of E is equal to 3 upon 6 which is equal to 1 upon 2. F consists of 2 elements so we get probability of F is equal to 2 upon 6 which is equal to 1 upon 3. Now E intersection F is the set consisting of elements which are common to both E and F and only element 3 is common to both E and F. So we get E intersection F is equal to singleton 3 and so probability of E intersection F is equal to 1 upon 6 because total number of elements are 6 out of which there is only one element in E intersection F. Now we mark this as 3, we mark this as 4 and we mark this as 5. So putting 4 and 5 in 1 we get probability of E given F is equal to 1 upon 6 upon 1 upon 3 and this is equal to 1 upon 2. Now putting 3 and 5 in 2 we get probability of F given E is equal to 1 upon 6 divided by 1 upon 2 and this is equal to 1 upon 3. Thus we have got our answer as 1 upon 2 comma 1 upon 3. Hope you have understood the solution. Bye and take care.