The numbers are selected at random (without replacement) from the first six positive integers. Let X denote the larger of the two numbers obtained. Find the probability distribution of the random variable X and hence find the means of the distribution.

Variable (X)23456
Probability P(X)115215315415515
First six numbers are 1,2,3,4,5,6.
X is bigger number among two numbers.
If X=2
for P(X)=probability of event that bigger of the two chosen number is 2.
So, Cases=(1,2)
So,P(X)=1/6C2=1/15—(1)
If X=3
So, favourable cases are=(1,3),(2,3)
P(X)=2/6C2=2/15—(2)
If X=4
So, favourable cases are=(1,4),(2,4),(3,4)
P(X)=3/6C2=3/15—(3)
If X=5
So, favourable cases are=(1,5),(2,5),(3,5),(4,5)
P(X)=4/6C2=4/15—(4)
If X=6
So, favourable cases are=(1,6),(2,6),(3,6),(4,6),(5,6)
P(X)=5/6C2=5/15—(5)
We can put all value of P(X),
Required mean =2 (1/15)+3 (2/15)+4 (3/15)+5 (4/15)+6 (5/15)=70/15⇒14/3