 Hello and welcome to the session I am Deepika here. Let's discuss the question which says which of the following arguments are correct and which are not correct? Give reasons for your answer. 1. If 2 points are continuously there are 3 possible outcomes 2 heads, 2 tails or 1 of each. Therefore for each of these outcomes the probability is 1 by 3. 2. If the guy is through there are 2 possible outcomes an odd number or an even number. Therefore the probability of getting an odd number is 1 by 2. Let us first understand the meaning of equally likely events. A given number of events to be equally likely tends to the other possibility of occurrence. So this is the key. We will take the help of this key idea to solve the above question. So let's start the solution. Continuously the possible outcomes are. Now we write H for head and T for tail. H that is head on the first coin and similarly head on the other coin. Tail on both the coins and tail on the first coin and head on the second coin and head on the first coin and the tail on the second coin. Therefore the number of possible outcomes is equal to 4 the event of getting both the coins. Therefore number of outcomes is equal to 1 that is the favorable outcomes for 2 heads is 1. Therefore the probability of the event E is equal to number of outcomes favorable to E upon total number of possible outcomes. So probability of getting 2 heads is 1 by 4. Note the event of getting 2 coins which are tossed simultaneously then probability of the event A is also 1 by 4. Let B remove the event 1 head and 1 tail. That is 1 of each when 2 coins are tossed simultaneously. Therefore the number of outcomes favorable to B is equal to 2. Hence probability of the event B is equal to 2 upon 4 which is equal to 1 by 2. Hence we have seen the probability of getting 2 heads when 2 coins are tossed simultaneously is 1 by 4. Similarly the probability of getting 2 tails when 2 coins are tossed simultaneously is 3 is 1 by 4. The probability of getting 1 of each when 2 coins are tossed simultaneously is 1 by 2. The probability of getting 1 of double the probability of getting 2 heads or 2 tails coins are tossed simultaneously. Hence our given statement is incorrect. Hence the given argument is incorrect. We can classify the outcomes like this. They are not equally likely. Hence the answer for the other question is our argument is incorrect. We can classify the outcomes like this that they are not then. Reason is that 1 of each can result in 2 heads from a 1st coin until the 2nd coin from a tail on the 1st coin and head on the 2nd coin. This makes it twice 3 as likely as 2 heads or 2 tails. So this is the answer for part 1. In part 2 a die is thrown. So when a die is thrown the possible outcomes are 1, 2, 3, 4, 5 and 6. Therefore the number of all possible outcomes is equal to 6. Let denote the event getting therefore probability of the event A is equal to now there are 3 even numbers. So this is equal to 3 upon 6 which is equal to 1 by 2. Let denote the event getting the odd number. Now there are 3 odd numbers 1, 3 and 5. Therefore the probability of the event B is equal to 3 upon 6 which is equal to 1 by 2. This event has the same possibility of occurrence in argument is correct because the 2 outcomes considering the question are equally likely. His answer for the above question is argument is correct. The 2 outcomes considering the question are equally likely. I hope the solution is clear to you. Bye and take care.