Question

Two numbers are selected at random (without replacement) the first six positive integers. Let $X$ denote the larger of the two numbers obtained. Find $E(X)$ .

Answer 1

Here in this type of problem, where we need to find $E(X)$ we need to know what the values of $X$ are. We need to find the total number of ways to select two numbers at random from the first six positive numbers. Then we need to find the individual probabilities of $X$ and then we need to apply the formula for $E(X) = \sum {{X_i}P({X_i})}$ and here ${X_i}$ is the value of the different values of $X$ and $P({X_i})$ represent the respective probability.

Complete step by step solution:
Here we are given that we need to select two numbers at random from the first six positive integers which are $1,2,3,4,5,6$
Now we need to understand how to calculate the number of ways to select two numbers from them. Let us consider an example: We have $26$ alphabets in English and we need to choose two out of them. So we know that the first alphabet can select from the $26$ alphabet and the second from the rest $25$ as we cannot replace the card. So we have a total of $(26 \times 25)$ ways.
Similarly here we have total of six numbers $1, 2, 3, 4, 5, 6$ so we can chose two numbers from them in $(6 \times 5) = 30{\text{ways}}$
Hence we can say that there are $30{\text{ ways}}$ to do so.
Now it is told that $X$ denotes the larger of the two numbers obtained. So if we have numbers $1, 2, 3, 4, 5, 6$ then from the two it is obvious that larger can never be $1$
Hence we get the value of $X = 2, 3, 4, 5, 6$
Now we need to calculate the probability when a larger number is $2, 3, 4, 5, 6$ individually.

1. When $X = 2$ possible cases can be $(1,2),(2,1)$
   So probability when $X = 2$ will be:
   $P(X = 2) = \dfrac{{{\text{number of ways when greater number is 2}}}}{{{\text{total ways to select two numbers}}}} = \dfrac{2}{{30}} = \dfrac{1}{{15}}$
2. When $X = 3$ possible cases can be $(1,3),(3,1),(2,3),(3,2)$
   So probability when $X = 3$ will be:
   $P(X = 3) = \dfrac{{{\text{number of ways when greater number is 3}}}}{{{\text{total ways to select two numbers}}}} = \dfrac{3}{{30}} = \dfrac{1}{{10}}$
3. When $X = 4$ possible cases can be $(1,4),(4,1),(2,4),(4,2),(3,4),(4,3)$
   So probability when $X = 4$ will be:
   $P(X = 4) = \dfrac{{{\text{number of ways when greater number is 4}}}}{{{\text{total ways to select two numbers}}}} = \dfrac{6}{{30}} = \dfrac{1}{5}$
4. When $X = 5$ possible cases can be $(1,5),(2,5),(3,5),(4,5),(5,1),(5,2),(5,3),(5,4)$
   So probability when $X = 5$ will be:
   $P(X = 5) = \dfrac{{{\text{number of ways when greater number is 5}}}}{{{\text{total ways to select two numbers}}}} = \dfrac{8}{{30}} = \dfrac{4}{{15}}$
5. When $X = 6$ possible cases can be $(1,6),(2,6),(3,6),(4,6),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5)$
   So probability when $X = 6$ will be:
   $P(X = 6) = \dfrac{{{\text{number of ways when greater number is 6}}}}{{{\text{total ways to select two numbers}}}} = \dfrac{{10}}{{30}} = \dfrac{1}{3}$
   So now we need to find the expected value of $X$ denoted by $E(X)$
   $E(X) = \sum {{X_i}P({X_i})}$ $= 2P(X = 2) + 3P(X = 3) + 4P(X = 4) + 5P(X = 5) + 6P(X = 6)$
   $E(X) = 2\left( {\dfrac{1}{{15}}} \right) + 3\left( {\dfrac{2}{{15}}} \right) + 4\left( {\dfrac{1}{5}} \right) + 5\left( {\dfrac{4}{{15}}} \right) + 6\left( {\dfrac{1}{3}} \right) = \dfrac{2}{{15}} + \dfrac{2}{5} + \dfrac{4}{5} + \dfrac{4}{3} + 2$
   So we get $E(X) = \dfrac{{2 + 6 + 12 + 20 + 30}}{{15}} = \dfrac{{70}}{{15}} = \dfrac{{14}}{3}$

So $E(X) = \dfrac{{14}}{3}$

Note:
Here a student can make a mistake by taking even $X = 1$ but it should not be taken as we just need to take those values of $X$ which are according to the given situation of the problem given. Hence we need to make the question clear first and then solve it carefully with proper calculations.