 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 the set which consists of elements 1, 3, 5. F is a set which consists of elements 2, 3. And G is a set which consists of elements 2, 3, 4, 5. Find probability of E given G and find probability of G 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 and here 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 elements 1, 2, 3, 4, 5 and 6. That is 6 elements. Now we are given that E is a set consisting of elements 1, 3, 5. F consists of 2, 3 and G is a set consisting of elements 2, 3, 4 and 5. And we have to find probability of E given G and probability of G given E. Now by key idea we know that probability of E given F is equal to probability of E intersection F upon probability of F. So probability of E given G is equal to probability of E intersection G upon probability of G and probability of G given E is equal to probability of G intersection E upon probability of E. We mark this as 1 and we mark this as 2. Now since E is a set consisting of 3 elements and S is a set consisting of 6 elements, So probability of E is equal to 3 upon 6 which is equal to 1 upon 2. Now G consists of 4 elements. So we have probability of G is equal to 4 upon 6 because total number of elements are 6 and this is equal to 2 upon 3. Now E intersection G is a set consisting of elements which are common to both E and G. Now here we can clearly see that 3 and 5 are the elements which are common to both E and G. So we have E intersection G is equal to a set which consists of elements 3 and 5. So we have probability of E intersection G is equal to 2 upon 6 which is equal to 1 upon 3. Now we mark this as 3, this as 4 and this as 5. So putting the value of 4 and 5 in 1 we get probability of E given G is equal to 1 upon 3 divided by 2 upon 3 and this is equal to 1 upon 2. And putting the values of 3 and 5 in 2 we get probability of G given E is equal to 1 upon 3 divided by 1 upon 2 and this is equal to 2 upon 3. So we have got our answer as 1 upon 2 comma 2 upon 3. Hope you have understood the solution. Bye and take care.