 Hello students, let's work out the following problem. It says a fair coin with one marked on one face and six on the other and A fair die are both tossed Find the probability that the sum of the numbers that turn up is number 1 3 number 2 12th, let us now move on to the solution And let us first write the sample space of this experiment. We are given a coin with one marked on one face and six on the other and With this coin we are tossing a die and we know that on a die we have six faces numbered 1 to 6 so the sample space That is the set of all possible outcomes It comes 1 1 that is we get 1 on the coin and 1 on the die the other possibility is that that we get 1 on the coin and 2 on the die Then 1 on the coin and 3 on the die 1 on the coin and 4 on the die 1 on the coin and 5 on the die 1 on the coin and 6 on the die now it may be possible that we get 6 on the coin and 1 on the die 6 on the coin 2 on the die Similarly other cases and we see that the total number of outcomes comes out to be 12 now in the first part we have to find the probability that the sum of the numbers is 1 So here e is the event of Getting sum of numbers Now we see that in how many cases we get the sum of numbers as 3 Now we see that when we get 1 on the coin and 2 on the die We get some as 3 and this is the only case when we get some as 3 so the number of outcomes Favorable to e is equal to 1 which is 1 2 so the probability that sum is 3 is The number of outcomes favorable to e which is 1 upon the total number of outcomes which are 12 Now in the second case we have to find the probability that the sum of the numbers is 12 So here e is the event Of getting sum of numbers as 12 Now we see that in how many cases we get the sum of numbers as 12 and we see that when we get 6 on the coin and 6 on the die we get some as 6 so the number of outcomes favorable to e Equal to 1 since there is only one such possibility Where we get some of the numbers as 12 which is 6 6 so the probability of Getting sum of numbers as 12 is The number of outcomes favorable to e that is 1 upon the total number of outcomes that is 12 so the probability of getting the sum of numbers as 12 is 1 by 12 So the probability of getting sum of numbers as 3 is one by twelve and the probability of getting sum of numbers as 12 is also 12 and We see that probability of getting some of numbers as 3 is is 1 by 12 and also sum of numbers is 12 is having probability 1 by 12. So this completes the question. Bye for now. Take care. Have a good day.