 Hello, and welcome to the session I am Deepika here. Let's discuss a question it says. In our claim, a man wins a rupee for a 6 and loses a rupee for any other number. When a fair guy is thrown, the man decided to throw a dice twice, but too quick as and when he gets a 6. Find the expected value of the amount he wins or loses. Now we know that the mean of x, denoted by mu, is the number sigma x i p r e taking from 1 to n. And it is also called the expectation of x, denoted by E s. In other words, the mean or expectation of a random variable x is the sum of the problems of all possible values of x by their respective probabilities. So this is a key idea behind our question. We will take up and pop this key idea to solve our question. So let's start the solution to the given question. In our game, a man wins a rupee for a 6 and loses a rupee. Let x take some values 1 and minus 1, 1 rupee, losing 1 rupee, getting us in a throw of the dice, equal to 1 over 6, is equal to 5 over 6. According to the given question, the man decided to throw a dice 3 for a 1 rupee 1. Now the equal to 1 over 6 x is equal to 1 over 6, which is equal to 0. Now we equal to 1 over 6 and probability of getting any other number apart from 6 is equal to 5 over 6 to 1 over 6, which is equal to 1 over 6, which is equal to 25 over 216. So the probability distribution of x, 1 over 6 when x is equal to 5 over 36 and when x is equal to minus 1, px is equal to 25 over 216. Now he wins or loses. Now here v is equal to sigma 36 plus n equal to 1 over over 216. Positive, therefore the expected value he wins is rupees 11 over 216. So the answer for the above question is 11 over 216. So this completes our session. I hope the solution is clear to you. Bye.