Question

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

Answer 1

The first six positive integers are: 1,2,3,4,5,6.

Let X be the larger number of two numbers selected, the possible outcomes are,

Sample space S is,

S={ ( 1,2 ),( 1,3 ),( 1,4 ),( 1,5 ),( 1,6 ),( 2,1 ),( 2,3 ),( 2,4 ),( 2,5 ),( 2,6 ), ( 3,1 ),( 3,2 ),( 3,4 ),( 3,5 ),( 3,6 ),( 4,1 ),( 4,2 ),( 4,3 ),( 4,5 ),( 4,6 ), ( 5,1 ),( 5,2 ),( 5,3 ),( 5,4 ),( 5,6 ),( 6,1 ),( 6,2 ),( 6,3 ),( 6,4 ),( 6,5 ) }

Here, S is the sample space.

Total number of possible outcomes is 30. The large number can be 2, 3, 4, 5 or 6.

Therefore the values of X can be 2, 3, 4, 5 or 6.

Consider X be the difference between the number, draw a table for the different sample space:

X	Events	Number of Outcomes	P( X )
2	{ ( 1,2 ),( 2,1 ) }	2	2 30
3	{ ( 3,1 ),( 3,2 ),( 2,3 ),( 1,3 ) }	4	4 30
4	{ ( 1,4 ),( 2,4 ),( 3,4 ),( 4,1 ),( 4,2 ),( 4,3 ) }	6	6 30
5	{ ( 1,5 ),( 2,5 ),( 3,5 ),( 4,5 ),( 5,1 ),( 5,2 ),( 5,3 ),( 5,4 ) }	8	8 30
6	{ ( 1,6 ),( 2,6 ),( 3,6 ),( 4,6 ),( 5,6 ),( 6,1 ),( 6,2 ),( 6,3 ),( 6,4 ),( 6,5 ) }	10	10 30

So the probability distribution:

X	2	3	4	5	6
P( X )	2 30	4 30	6 30	8 30	10 30

Therefore, the mean number is given by,

μ=E( X ) = ∑ i=1 n X i P i =( 2× 2 30 )+( 3× 4 30 )+( 4× 6 30 )+( 5× 8 30 )+( 6× 10 30 ) = 140 30 = 14 3

Thus, the mean number is 14 3 .