 Good morning friends, I am Purva and today we will discuss the following question. A fair die is ruled. Consider events E which is equal to a set consisting of elements 1, 3, 5. F is a set consisting of elements 2, 3 and G is a set consisting of elements 2, 3, 4, 5. Then find probability of E union F given G and probability of E intersection F given G. 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. Now when we roll a die we either get a 1 or 2 or 3 or 4 or 5 or 6. So here we get S is equal to a set which consists of elements 1, 2, 3, 4, 5, 6. That is F consists of 6 elements. Now we are given that E consists of elements 1, 3, 5, F consists of 2, 3 and G consists of 2, 3, 4 and 5. We have to find probability of E union F given G and probability of E intersection F given G. Now by key idea we know that if E and F are two events then probability of E given F is equal to probability of E intersection F upon probability of F. So here probability of E union F given G is equal to probability of E union F intersection G upon probability of G. And probability of E intersection F given G is equal to probability of E intersection F intersection G upon probability of G. We mark this as 1 and we mark this as 2. Now probability of G is equal to 4 upon 6 because G is a set consisting of 4 elements whereas in all we have 6 elements and this is equal to 2 upon 3. We mark this as 3. E union F is a set which consists of elements which are either in E or in F. Now here we can clearly see that 1, 2, 3 and 5 are the elements which are either in E or in F. So we get E union F is equal to a set which consists of elements 1, 2, 3 and 5. E union F intersection G is a set which consists of elements which are both in E union F and in G. Now E union F consists of elements 1, 2, 3 and 5 whereas G consists of elements 2, 3, 4 and 5. So E union F intersection G consists of 2, 3 and 5. That is the elements which are common to both G and E union F. So we get probability of E union F intersection G is equal to 3 upon 6 and this is equal to 1 upon 2. E intersection F is a set which consists of elements which are both in E and in F. Now here we can clearly see that only element 3 is common to both E and F. So we get E intersection F is equal to singleton 3. E intersection F intersection G is equal to a set which consists of elements which are common to both E intersection F and 2G. Now E intersection F consists of only one element that is 3 and G also consists of that element. So we get E intersection F intersection G is equal to singleton 3. So probability of E intersection F intersection G is equal to 1 upon 6 because E intersection F intersection G consists of only one element out of the six elements. Now we mark this as 4 and we mark this as 5. Now putting 3 and 4 in 1 we get probability of E union F given G is equal to 1 upon 2 divided by 2 upon 3 and this is equal to 3 upon 4. Now putting 3 and 5 in 2 we get probability of E intersection F given G is equal to 1 upon 6 divided by 2 upon 3 and this is equal to 1 upon 4. So we have got our answer as 3 upon 4 comma 1 upon 4. Hope you have understood the solution. Bye and take care.