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

Toolbox:
• Mean of the probability distribution = $\sum (X_i \times P(X_i))$ The Expected value of X is nothing but the mean of X.
The first six positive integers are 1, 2, 3, 4, 5, 6.
We can select the two positive numbers in 6 $\times$ 5 = 30 different ways.
Out of this, 2 numbers are selected at random and let X denote the larger of the two numbers.
Since X is the large of the two numbers, X can assume the value of 2, 3, 4, 5 or 6.
P (X =2) = P (larger number is 2) = {(1,2) and (2,1)} = $\large\frac{2}{30}$
P (X = 3) = P (larger number is 3) = {(1,3), (3,1), (2,3), (3,2)} = $\large\frac{4}{30}$
P (X = 4) = P (larger number is 4) = {(1,4), (4,1), (2,4), (4,2), (3,4), (4,3)} = $\large\frac{6}{30}$
P (X = 5) = P (larger number is 5) = {(1,5), (51,), (2,5), (5,2), (3,5), (5,3), (4,5), (5.4)} = $\large\frac{8}{30}$
P (X = 6) = P (larger number is 6) = {(1,6), (6,1), (2,6), (6,2), (3,6), (6,3), (4,6), (6,4), (5,6), (6,5)} = $\large\frac{10}{30}$
Given the above probability distribution, the expected value or the mean can be calculated as follows:
Mean = $\sum (X_i \times P(X_i))$ = $2 \times \large\frac{2}{30} + $$3\times\large\frac{4}{30} +$$4\times\large\frac{6}{30} +$$5\times\large\frac{8}{30}+$$6\times\large\frac{10}{30} = \frac{4+12+24+40+60}{30} = \frac{140}{30} = \frac{14}{3}$