 Good morning friends. I am Pooja and today we will work out the following question. Given that the two numbers appearing on drawing two dice are different, find the probability of the event, the sum of numbers on the dice is 4. Now if E and F are two events associated with the sample of a random experiment, 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 tossing two dice. Now we know that a die is a cube having six faces. Now when we toss a die, any of the numbers from 1 to 6 can appear on the top face. So on tossing one die we can get six outcomes. Now if we toss two die and if we get one on the first die and one on the second die, then we get the outcome as 1, 1. Similarly if we get one on the first die and two on the second die, we get the outcome as 1, 2. So we can get all these outcomes on tossing two dice, that is we can get total of 36 outcomes on tossing two dice. Now we have suppose that S is the sample space of experiment of tossing two dice. So we get S is equal to S is a set consisting of 1, 1, 1, 2, 1, 3, 1, 4 and so on till 6, 6 that is it consists of all these outcomes. Now the sample space has 36 outcomes. Now in the question we are given that the two numbers appearing on throwing two dice are different and we have to find the probability of the event that the sum of numbers on the dice is 4. So let E be the event that sum of numbers on the dice is 4 and let F be the event that two numbers appearing are different. Now here we can clearly see that out of all these 36 outcomes the outcomes 1, 3, 2, 2 and 3, 1 these are the three outcomes that give us the sum of numbers on the dice as 4. For all other outcomes we get the sum of numbers on the dice as either less than 4 or greater than 4. In case of outcome 1, 3 we can clearly see that 1 plus 3 is equal to 4. Here we get 2 plus 2 is equal to 4 and here we get 3 plus 1 is equal to 4. So we get E as the set consisting of these three outcomes that is 1, 3, 2, 2 and 3, 1. Now F is the event that two numbers appearing on throwing of two dice are different. Now leaving aside the outcomes that appear on this diagonal all other outcomes consist of two numbers that on throwing of two dice are different. So we get F to be a set consisting of 1, 2, 1, 3, 1, 4, 1, 5, 1, 6, 2, 1, 2, 3 and so on. That is 36 minus 6 which is equal to 30 outcomes. Now probability of F is equal to 30 upon 36. Because total number of outcomes are 36 and F has 30 outcomes and this is equal to 15 upon 18. E intersection F is equal to all those outcomes which are common to both E and F. Now E consists of only three outcomes 1, 3, 2, 2 and 3, 1 out of which only 1, 3 and 3, 1 consists of two numbers which are different on tossing of two dice. So we get E intersection F is equal to a set which consists of 1, 3 and 3, 1. So we get probability of E intersection F is equal to 2 upon 36 and this is equal to 1 upon 18. Now we have to find the probability of the event that the sum of numbers on the dice is 4 given that the two numbers appearing on throwing two dice are different. So we have to find probability of E given F. 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 we have probability of E given F is equal to probability of E intersection F upon probability of F and this is equal to now probability of E intersection F is equal to 1 upon 18 upon probability of F is equal to 15 upon 18. And this is equal to 1 upon 15. So we have got our answer as 1 upon 15. Hope you have understood the solution. Bye and take care.