 Hello and welcome to the session. I am Deepika here. Let's discuss the question which says refer to example 13 Example 13 says two dice one view and one break are thrown at the same time Write down all the possible outcomes What is the probability that the sum of the two numbers appearing on the top of the dice is? number 1 8 number 2 13 number 3 less than or equal to 10 Then complete the following table now if the sum of two dice is 2 then probability is 1 over 36 and if someone two dice is 8 probability is 5 over 36 and if someone two dice is 12 Probability is 1 over 36 R2 a student argues that there are 11 possible outcomes 2 3 4 5 6 7 8 9 10 11 and 12 Therefore each of them has a probability 1 over 11. Do you agree with this argument? justify your answer Now we know that the probability of an event e is written as p equal to number of outcomes Acible outcomes of the experiment. So this is a key idea if you find a question We will take the help of this key idea to solve the above question. So let's start the solution Now when two dice are thrown continuously one first time and one on the second 1 on the first die and 2 on the second die 1 on the 1st die and 4 on the 2nd die 1 on the 1st die and 5 on the 2nd die and again 1 on the 1st die are 6 on the 2nd die And 1 to 6 on the second guy, so the ordered pairs are 2, 1 and 1, 2, 6 on the second guy. So the ordered pairs are first guy and 1 to 6 first guy and 1 to 6 ordered pairs are 5, 1 first guy and 1 to 6, so the ordered pairs are 6, 1, 6, 3, 6, 4 possible outcomes equal to 6 into 6 which is equal to 36. In the event 2 dies a table, the outcomes favorable to 1 to the event, the number of outcomes favorable to the event is equal to 2. Hence probability of the event A is equal to 2 upon 36. Let's be the event of getting the sum on. So from this table, the outcomes are 1, 3, so the number of outcomes favorable to the event probability of the event B is equal to 3 upon 36. Getting the sum on 2 dies, so from this table the outcomes favorable to the event are 1, 4, that is 1, 2, 3, 4, these are the number of outcomes favorable to the event C. Hence probability of the event C 4 upon 36, let be the event of the outcomes favorable to the event DR, 5, that is the number of outcomes favorable to the event DR, 1, 2, the probability of the event D is equal to getting the sum on 2 dies as 7, so from this table the outcomes favorable to the event E are 1, 6. Hence the number of outcomes favorable to the event E, probability of the event E is equal to 6 upon 36. Let be the event of the sum on favorable to the event F. So the number of outcomes favorable to the event F is equal to 4. Hence probability of the event F is equal to 4 over 36. Let be the event of getting the sum on 2 dies, go to the event G. The number of outcomes favorable to the event G is equal to 3, hence the probability of the event G on 36. The event of getting the sum on 2 dies favorable to the event HR, so the probability of the event H is equal to 2 upon 36. Therefore, the required solution is, now if the sum on 2 dies as 3 its probability is 2 over 36 and if the sum is 4 its probability is 3 over 36 and for the sum 5 probability is 4 over 36 and if the sum on 2 dies as 6 its probability is 5 over 36 and for the sum 7 it is 6 over 36 and for the sum 9 it is 4 over 36 and for 10 it is 3 over 36 for 11 it is 2 over 36. So this is the answer for part 1. Let's move to the part, argues that there are 11 possible outcomes 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 and 12. Therefore, each of them has a probability 1 by 11. To observing this table, we do not agree this argument not equally likely. We do not agree with the students argument the 11 sums likely. I hope the solution is clear to you. Bye and take care.