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 .