 Good morning friends. I am Purva and today we will discuss the following question find the main number of heads in three tosses of a fair coin Now let X be a random variable whose possible values X1 X2 up till Xn are covered probabilities P1 P2 up till Pn respectively Then the mean of random variable X is also called the expectation of X Which is denoted by e of X and this is equal to mu as the mean of X is also denoted by mu and this is equal to summation i is equal to 1 to n Xi pi So this is the key idea behind our question. Let us begin with the solution now Now let S be the sample space For three tosses of a fair coin Then S is a set which consists of eight outcomes. That is head tail head head head head tail tail tail tail tail head head head tail tail head tail head tail tail and tail head head Now since we have to find the mean number of heads in three tosses of a fair coin So let X denotes the number of heads in three tosses of a fair coin Then we have X is equal to 0 1 2 3 Because the maximum number of heads in three tosses of a coin is three and the minimum number of heads is zero That is when all our tails Now we find the probability distribution of X. So the probability distribution of X is as follows When X is equal to zero that is the number of heads in three tosses of a coin is zero That is when all our tails then we have only one outcome out of the eight outcomes, which is favorable So we get probability of X as 1 upon 8 When X is equal to 1 that is when we have only one head and two tails Then we have three favorable outcomes out of these eight outcomes So we get probability of X as 3 upon 8 When X is equal to 2 that is number of heads is 2 and number of tail is 1 then also we have Three possible outcomes out of these eight outcomes. So we get probability of X as 3 upon 8 When X is equal to 3 that is number of heads is equal to 3 That is when all of them are head then there is only one possible outcome and the total number of outcomes is 8 So we get probability of X as 1 upon 8 So we have got the probability distribution of X Now we will find the mean of this random variable X Now by key idea we know that the mean of a random variable X is given by Expectation of X which is equal to summation i is equal to 1 to n X i p i So here we have mean of random variable X is Also called expectation of X and this is given by summation i is equal to 1 to n X i into p i Now here the possible values of X are 0 1 2 3 so these are our X i's And the probabilities are 1 upon 8 3 upon 8 3 upon 8 1 upon 8 respectively So these are our p i's So we have expectation of X is equal to X 1 p 1 that is 0 into 1 upon 8 As here X 1 is equal to 0 and p 1 is equal to 1 upon 8 plus X 2 p 2 that is 1 into 3 upon 8 Plus X 3 p 3 that is X 3 is equal to 2 and p 3 is equal to 3 upon 8 So we get 2 into 3 upon 8 Plus X 4 p 4 that is 3 into 1 upon 8 This is equal to 0 into 1 upon 8 is 0 plus 1 into 3 upon 8 is 3 upon 8 Plus 2 into 3 upon 8 is 6 upon 8 plus 3 into 1 upon 8 is 3 upon 8 This is equal to 12 upon 8 Which is equal to 3 upon 2 and This is further equal to 1 point 5 So the mean number of heads in three tosses of a fair coin is 1 point 5 This is our answer. Hope you have understood the solution. Bye and take care