Question 1025315
Look at the possible 30 outcomes,
(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)
Then the max value of the pair,
2 3 4 5 6
2 3 4 5 6
3 3 4 5 6
4 4 4 5 6
5 5 5 5 6
6 6 6 6 6
The number of occurences of each number,
1=>0
2=>2
3=>4
4=>6
5=>8
6=>10
So then the probability distirubtion is,
{{{P[1]=0/30=0}}}
{{{P[2]=2/30=1/15}}}
{{{P[3]=4/30=2/15}}}
{{{P[4]=6/30=1/5}}}
{{{P[5]=8/30=4/15}}}
{{{P[6]=10/30=1/3}}}